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.util.*;
  import java.io.*;
  import java.text.*;
  
  public class CF_1552_C{
      //SOLUTION BEGIN
      void pre() throws Exception{}
      void solve(int TC) throws Exception{
          int N = ni(), K = ni();
          int[][] seg = new int[N][];
          boolean[] used = new boolean[2*N];
          for(int i = 0; i< K; i++){
              int u = ni()-1, v = ni()-1;
              used[u] = used[v] = true;
              seg[i] = new int[]{Math.min(u, v), Math.max(u, v)};
          }
          List<Integer> list = new ArrayList<>();
          for(int i = 0; i< 2*N; i++){
              if(used[i])continue;
              list.add(i);
          }
          for(int j = K, i = 0; j< N; j++, i++)
              seg[j] = new int[]{list.get(i), list.get(i+N-K)};
  
  
          Arrays.sort(seg, (int[] i1, int[] i2) -> Integer.compare(i1[0], i2[0]));
          int ans = 0;
          for(int i = 0; i< N; i++){
              for(int j = i+1; j< N; j++){
                  if(seg[i][1] > seg[j][0] && seg[i][1] < seg[j][1])ans++;
              }
          }
          pn(ans);
  
      }
      //SOLUTION END
      void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
      void exit(boolean b){if(!b)System.exit(0);}
      static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
      final long IINF = (long)1e17;
      final int INF = (int)1e9+2;
      DecimalFormat df = new DecimalFormat("0.00000000000");
      double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-8;
      static boolean multipleTC = true, memory = true, fileIO = false;
      FastReader in;PrintWriter out;
      void run() throws Exception{
          long ct = System.currentTimeMillis();
          if (fileIO) {
              in = new FastReader("");
              out = new PrintWriter("");
          } else {
              in = new FastReader();
              out = new PrintWriter(System.out);
          }
          //Solution Credits: Taranpreet Singh
          int T = multipleTC? ni():1;
          pre();
          for (int t = 1; t <= T; t++) solve(t);
          out.flush();
          out.close();
          System.err.println(System.currentTimeMillis() - ct);
      }
      public static void main(String[] args) throws Exception{
          if(memory)new Thread(null, new Runnable() {public void run(){try{new CF_1552_C().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 28).start();
          else new CF_1552_C().run();
      }
      int[][] make(int n, int e, int[] from, int[] to, boolean f){
          int[][] g = new int[n][];int[]cnt = new int[n];
          for(int i = 0; i< e; i++){
              cnt[from[i]]++;
              if(f)cnt[to[i]]++;
          }
          for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
          for(int i = 0; i< e; i++){
              g[from[i]][--cnt[from[i]]] = to[i];
              if(f)g[to[i]][--cnt[to[i]]] = from[i];
          }
          return g;
      }
      int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
          int[][][] g = new int[n][][];int[]cnt = new int[n];
          for(int i = 0; i< e; i++){
              cnt[from[i]]++;
              if(f)cnt[to[i]]++;
          }
          for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
          for(int i = 0; i< e; i++){
              g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
              if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
          }
          return g;
      }
      int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
      int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
      long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
      int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
      int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
      void p(Object... o){for(Object oo:o)out.print(oo+" ");}
      void pn(Object... o){for(int i = 0; i< o.length; i++)out.print(o[i]+(i+1 < o.length?" ":"\n"));}
      void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
      String n()throws Exception{return in.next();}
      String nln()throws Exception{return in.nextLine();}
      int ni()throws Exception{return Integer.parseInt(in.next());}
      long nl()throws Exception{return Long.parseLong(in.next());}
      double nd()throws Exception{return Double.parseDouble(in.next());}
  
      class FastReader{
          BufferedReader br;
          StringTokenizer st;
          public FastReader(){
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          public FastReader(String s) throws Exception{
              br = new BufferedReader(new FileReader(s));
          }
  
          String next() throws Exception{
              while (st == null || !st.hasMoreElements()){
                  try{
                      st = new StringTokenizer(br.readLine());
                  }catch (IOException  e){
                      throw new Exception(e.toString());
                  }
              }
              return st.nextToken();
          }
  
          String nextLine() throws Exception{
              String str;
              try{
                  str = br.readLine();
              }catch (IOException e){
                  throw new Exception(e.toString());
              }
              return str;
          }
      }
  }