description: On a circle lie 2n distinct points, with the following property: however you choose 3 chords that connect 3 disjoint pairs of points, no point strictly inside the circle belongs to all 3 chords. The points are numbered 1,   2,   ...,   2n in clockwise order.
  
  Initially, k chords connect k pairs of points, in such a way that all the 2k endpoints of these chords are distinct.
  
  You want to draw n - k additional chords that connect the remaining 2(n - k) points (each point must be an endpoint of exactly one chord).
  
  In the end, let x be the total number of intersections among all n chords. Compute the maximum value that x can attain if you choose the n - k chords optimally.
  
  Note that the exact position of the 2n points is not relevant, as long as the property stated in the first paragraph holds.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
  
  The first line of each test case contains two integers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ n) — half the number of points and the number of chords initially drawn.
  
  Then k lines follow. The i-th of them contains two integers x_i and y_i (1 ≤ x_i,   y_i ≤ 2n, x_i ≠ y_i) — the endpoints of the i-th chord. It is guaranteed that the 2k numbers x_1,   y_1,   x_2,   y_2,   ...,   x_k,   y_k are all distinct.
  
  Output
  
  For each test case, output the maximum number of intersections that can be obtained by drawing n - k additional chords.
  
  Example
  
  Input
  
  
  4
  4 2
  8 2
  1 5
  1 1
  2 1
  2 0
  10 6
  14 6
  2 20
  9 10
  13 18
  15 12
  11 7
  
  
  Output
  
  
  4
  0
  1
  14
  
  Note
  
  In the first test case, there are three ways to draw the 2 additional chords, shown below (black chords are the ones initially drawn, while red chords are the new ones):
  
  <image>
  
  We see that the third way gives the maximum number of intersections, namely 4.
  
  In the second test case, there are no more chords to draw. Of course, with only one chord present there are no intersections.
  
  In the third test case, we can make at most one intersection by drawing chords 1-3 and 2-4, as shown below:
  
  <image>
solution: import java.io.*;
  import java.util.*;
  import java.math.BigInteger;
  
  /**
   __                  __
   ( _)                ( _)
   / / \\              / /\_\_
   / /   \\            / / | \ \
   / /     \\          / /  |\ \ \
   /  /   ,  \ ,       / /   /|  \ \
   /  /    |\_ /|      / /   / \   \_\
   /  /  |\/ _ '_| \   / /   /   \    \\
   |  /   |/  0 \0\    / |    |    \    \\
   |    |\|      \_\_ /  /    |     \    \\
   |  | |/    \.\ o\o)  /      \     |    \\
   \    |     /\\`v-v  /        |    |     \\
   | \/    /_| \\_|  /         |    | \    \\
   | |    /__/_ `-` /   _____  |    |  \    \\
   \|    [__]  \_/  |_________  \   |   \    ()
   /    [___] (    \         \  |\ |   |   //
   |    [___]                  |\| \|   /  |/
   /|    [____]                  \  |/\ / / ||
   (  \   [____ /     ) _\      \  \    \| | ||
   \  \  [_____|    / /     __/    \   / / //
   |   \ [_____/   / /        \    |   \/ //
   |   /  '----|   /=\____   _/    |   / //
   __ /  /        |  /   ___/  _/\    \  | ||
   (/-(/-\)       /   \  (/\/\)/  |    /  | /
   (/\/\)           /   /   //
   _________/   /    /
   \____________/    (
  
  
   @author NTUDragons-Reborn
   */
  
  public class Solution{
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          Task solver = new Task();
          solver.solve(in, out);
          out.close();
      }
      // main solver
      static class Task{
  
          double eps= 0.00000001;
          static final int MAXN = 1010;
          static final int MOD= 1000000007;
  
          // stores smallest prime factor for every number
          static int spf[] = new int[MAXN];
  
          static boolean[] prime;
  
          Map<Integer,Set<Integer>> dp= new HashMap<>();
  
          // Calculating SPF (Smallest Prime Factor) for every
          // number till MAXN.
          // Time Complexity : O(nloglogn)
          public void sieve()
          {
              spf[1] = 1;
              for (int i=2; i<MAXN; i++)
  
                  // marking smallest prime factor for every
                  // number to be itself.
                  spf[i] = i;
  
              // separately marking spf for every even
              // number as 2
              for (int i=4; i<MAXN; i+=2)
                  spf[i] = 2;
  
              for (int i=3; i*i<MAXN; i++)
              {
                  // checking if i is prime
                  if (spf[i] == i)
                  {
                      // marking SPF for all numbers divisible by i
                      for (int j=i*i; j<MAXN; j+=i)
  
                          // marking spf[j] if it is not
                          // previously marked
                          if (spf[j]==j)
                              spf[j] = i;
                  }
              }
          }
  
          void sieveOfEratosthenes(int n)
          {
              // Create a boolean array
              // "prime[0..n]" and
              // initialize all entries
              // it as true. A value in
              // prime[i] will finally be
              // false if i is Not a
              // prime, else true.
              prime= new boolean[n + 1];
              for (int i = 0; i <= n; i++)
                  prime[i] = true;
      
              for (int p = 2; p * p <= n; p++)
              {
                  // If prime[p] is not changed, then it is a
                  // prime
                  if (prime[p] == true)
                  {
                      // Update all multiples of p
                      for (int i = p * p; i <= n; i += p)
                          prime[i] = false;
                  }
              }
          }
  
          // A O(log n) function returning primefactorization
          // by dividing by smallest prime factor at every step
          public Set<Integer> getFactorization(int x)
          {
              if(dp.containsKey(x)) return dp.get(x);
              Set<Integer> ret = new HashSet<>();
              while (x != 1)
              {
                  if(spf[x]!=2) ret.add(spf[x]);
                  x = x / spf[x];
              }
              dp.put(x,ret);
              return ret;
          }
          // function to find first index >= x
          public int lowerIndex(List<Integer> arr, int n, int x)
          {
              int l = 0, h = n - 1;
              while (l <= h)
              {
                  int mid = (l + h) / 2;
                  if (arr.get(mid) >= x)
                      h = mid - 1;
                  else
                      l = mid + 1;
              }
              return l;
          }
          public int lowerIndex(int[] arr, int n, int x)
          {
              int l = 0, h = n - 1;
              while (l <= h)
              {
                  int mid = (l + h) / 2;
                  if (arr[mid] >= x)
                      h = mid - 1;
                  else
                      l = mid + 1;
              }
              return l;
          }
  
          // function to find last index <= y
          public int upperIndex(List<Integer> arr, int n, int y)
          {
              int l = 0, h = n - 1;
              while (l <= h)
              {
                  int mid = (l + h) / 2;
                  if (arr.get(mid) <= y)
                      l = mid + 1;
                  else
                      h = mid - 1;
              }
              return h;
          }
          public int upperIndex(int[] arr, int n, int y)
          {
              int l = 0, h = n - 1;
              while (l <= h)
              {
                  int mid = (l + h) / 2;
                  if (arr[mid] <= y)
                      l = mid + 1;
                  else
                      h = mid - 1;
              }
              return h;
          }
  
          // function to count elements within given range
          public int countInRange(List<Integer> arr, int n, int x, int y)
          {
              // initialize result
              int count = 0;
              count = upperIndex(arr, n, y) -
                      lowerIndex(arr, n, x) + 1;
              return count;
          }
  
          public int _gcd(int a,  int  b)
          {
  
              if(b == 0) {
                  return a;
              }
              else {
                  return _gcd(b, a % b);
              }
          }
  
          public int add(int a, int b){
              a+=b;
              if(a>=MOD) a-=MOD;
              else if(a<0) a+=MOD;
              return a;
          }
  
          public int mul(int a, int b){
              long res= (long)a*(long)b;
              return (int)(res%MOD);
          }
  
          public int power(int a, int b) {
              int ans=1;
              while(b>0){
                  if((b&1)!=0) ans= mul(ans,a);
                  b>>=1;
                  a= mul(a,a);
              }
              return ans;
          }
  
          int[] fact= new int[MAXN];
          int[] inv= new int[MAXN];
  
          public int Ckn(int n, int k){
              if(k<0 || n<0) return 0;
              return mul(mul(fact[n],inv[k]),inv[n-k]);
          }
  
          public int inverse(int a){
              return power(a,MOD-2);
          }
  
          public void preprocess() {
              fact[0]=1;
              for(int i=1;i<MAXN;i++) fact[i]= mul(fact[i-1],i);
              inv[MAXN-1]= inverse(fact[MAXN-1]);
              for(int i=MAXN-2;i>=0;i--){
                  inv[i]= mul(inv[i+1],i+1);
              }
          }
  
          /**
           * return VALUE of lower bound for unsorted array
           */
          public int lowerBoundNormalArray(int[] arr, int x){
              TreeSet<Integer> set= new TreeSet<>();
              for(int num: arr) set.add(num);
              return set.lower(x);
          }
          /**
           * return VALUE of upper bound for unsorted array
           */
          public int upperBoundNormalArray(int[] arr, int x){
              TreeSet<Integer> set= new TreeSet<>();
              for(int num: arr) set.add(num);
              return set.higher(x);
          }
  
          public void debugArr(int[] arr){
              for(int i: arr) out.print(i+" ");
              out.println();
          }
  
          public int rand(){
              int min=0, max= MAXN;
              int random_int = (int)Math.floor(Math.random()*(max-min+1)+min);
              return random_int;
          }
          
          InputReader in; PrintWriter out;
          
          public void solve(InputReader in, PrintWriter out) {
              this.in=in; this.out=out;
              int t= in.nextInt();
              while(t-->0){
                  solveC();
              }
          }
          public void solveA(){
              int n= in.nextInt();
              char[] s= in.nextStringCharArr();
              int[] a = new int[n];
              int[] b = new int[n];
              for (int i = 0; i < n ;i++){
                  a[i] = b[i] = s[i] - 'a';
              }
              Arrays.sort(b);
              int ans=0;
              for(int i = 0; i < n; i++){
                  ans+=(a[i]!=b[i]?1:0);
              }
              out.println(ans);
          }
          public void solveB(){
              int n= in.nextInt();
              int[][] A= new int[n+1][5];
              for (int i = 1; i <= n; i++){
                  for (int j = 0; j < 5; j++){
                      A[i][j]= in.nextInt();
                  }
              }
              int candi = 1;
              for (int i = 2; i <= n; i++){
                  int cnt = 0;
                  for (int j = 0; j < 5; j++){
                      cnt+=(A[candi][j] < A[i][j]?1:0);
                  }
                  if (cnt < 3) candi = i;
              }
              boolean ok = true;
              for (int i = 1; i <= n; i++){
                  if (i == candi) continue;
                  int cnt = 0;
                  for (int j = 0; j < 5; j++){
                      cnt+=(A[candi][j] < A[i][j]?1:0);
                  }
                  if (cnt < 3){
                      ok  = false; break;
                  }
              }
              if(ok) out.println(candi);
              else out.println(-1);
          }
          public void solveC(){
              int n= in.nextInt(), k= in.nextInt();
              boolean[] vis= new boolean[2*n+1];
              List<Pair> segs= new ArrayList<>();
              for (int i = 1; i <= k; i++){
                  int u=in.nextInt(), v= in.nextInt();
                  vis[u] = vis[v] = true;
                  if (u > v) {
                      int temp=u;
                      u=v;
                      v= temp;
                  }
                  segs.add(new Pair(u, v));
              }
              List<Integer> leftOver= new ArrayList<>();
              for (int i = 1; i <= 2 * n; i++){
                  if (!vis[i]){
                      leftOver.add(i);
                  }
              }
              for (int i = 0; i < n - k; i++){
                  segs.add(new Pair(leftOver.get(i), leftOver.get(i+n-k)));
              }
              int ans = 0;
              for (int i = 0; i < n; i++){
                  for (int j = i + 1; j < n; j++){
                      int l1 = segs.get(i).first, r1 = segs.get(i).second;
                      int l2 = segs.get(j).first, r2 = segs.get(j).second;
                      boolean ok1 = (l1 < l2 && l2 < r1);
                      boolean ok2 = (l1 < r2 && r2 < r1);
                      if (ok1 && ok2){
                          continue;
                      }
                      if (!ok1 && !ok2){
                          continue;
                      }
                      ans++;
                  }
              }
              out.println(ans);
            
          }
  
      }
  
      static class Venice{
          public Map<Long,Long> m= new HashMap<>();
          public long base=0;
          public long totalValue=0;
          private int M= 1000000007;
  
          private long addMod(long a, long b){
              a+=b;
              if(a>=M) a-=M;
              return a;
          }
  
          public void reset(){
              m= new HashMap<>();
              base=0;
              totalValue=0;
          }
          public void update(long add){
              base= base+ add;
          }
          public void add(long key, long val){
              long newKey= key-base;
              m.put(newKey, addMod(m.getOrDefault(newKey,(long)0),val));
          }
      }
  
  
  
      static class Tuple implements Comparable<Tuple>{
          int x, y, z;
          public Tuple(int x, int y, int z){
              this.x= x;
              this.y= y;
              this.z=z;
          }
          @Override
          public int compareTo(Tuple o){
              return this.x-o.x;
          }
      }
  
      static class Pair implements Comparable<Pair>{
          public int first;
          public int second;
          public Pair(int x, int y){
              this.first= x;
              this.second= y;
          }
  
          @Override
          public int compareTo(Pair o) {
              return this.first-o.first;
          }
      }
  
      // public static class compareL implements Comparator<Tuple>{
      //     @Override
      //     public int compare(Tuple t1, Tuple t2) {
      //         return t2.l - t1.l;
      //     }
      // }
  
      // fast input reader class;
      static class InputReader {
          BufferedReader br;
          StringTokenizer st;
  
          public InputReader(InputStream stream) {
              br = new BufferedReader(new InputStreamReader(stream));
          }
  
          public String nextToken() {
              while (st == null || !st.hasMoreTokens()) {
                  String line = null;
                  try {
                      line = br.readLine();
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
                  if (line == null) {
                      return null;
                  }
                  st = new StringTokenizer(line);
              }
              return st.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(nextToken());
          }
          public double nextDouble(){
              return Double.parseDouble(nextToken());
          }
          public long nextLong(){
              return Long.parseLong(nextToken());
          }
          public int[] nextIntArr(int n){
              int[] arr= new int[n];
              for(int i=0;i<n;i++) arr[i]= nextInt();
              return arr;
          }
          public long[] nextLongArr(int n){
              long[] arr= new long[n];
              for(int i=0;i<n;i++) arr[i]= nextLong();
              return arr;
          }
          public List<Integer> nextIntList(int n){
              List<Integer> arr= new ArrayList<>();
              for(int i=0;i<n;i++) arr.add(nextInt());
              return arr;
          }
          public int[][] nextIntMatArr(int n, int m){
              int[][] mat= new int[n][m];
              for(int i=0;i<n;i++) for(int j=0;j<m;j++) mat[i][j]= nextInt();
              return mat;
          }
          public List<List<Integer>> nextIntMatList(int n, int m){
              List<List<Integer>> mat= new ArrayList<>();
              for(int i=0;i<n;i++){
                  List<Integer> temp= new ArrayList<>();
                  for(int j=0;j<m;j++) temp.add(nextInt());
                  mat.add(temp);
              }
              return mat;
          }
          public char[] nextStringCharArr(){
              return nextToken().toCharArray();
          }
      }
  }