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 static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import java.util.*;
  import java.io.*;
  
  public class A
  {
      static int n, k;
      static int[] arr;
      static char[] s;
      public static void main(String[] args) throws IOException
      {
          f = new Flash();
          out = new PrintWriter(System.out);
          
          int T = ni();
          for(int tc = 1; tc <= T; tc++){
          	n = ni(); k = ni(); fn();
          }
          
          out.flush(); out.close();
      }
      
      static void fn()
      {
          int[] arr = new int[2*n+1];
          List<int[]> A = new ArrayList<>();
          int cnt = 0;
          for(int i = 1; i <= k; i++) {
          	int[] a = new int[2];
          	a[0] = ni(); a[1] = ni(); sort(a);
          	A.add(new int[] {i, a[0], a[1]});
          	arr[a[0]] = i; arr[a[1]] = -i;
          	cnt++;
          }
          
          for(int i = cnt+1; i <= n; i++) {
          	int l = 0, r = 0, space = n-i;
          	for(int j = 1; j <= 2*n; j++) {
          		if(arr[j] != 0) continue;
          		arr[j] = i; l = j;
          		break;
          	}
          	
          	for(int j = 2*n; j >= 1; j--) {
          		if(arr[j] != 0) continue;
          		if(space == 0) {
          			arr[j] = -i; r = j;
          			break;
          		}
          		
          		space -= 1;
          	}
          	
          	A.add(new int[] {i, l, r});
          }
          
  //        debug(arr);
          long ans = 0;
          for(int[] p1 : A) {
          	for(int[] p2 : A) {
          		if(p1[0] == p2[0]) continue;
          		if(p1[1] < p2[1] && p2[1] < p1[2] && p1[2] < p2[2]) ans++;
          	}
          }
          
          sop(ans);
      }
      
      static Flash f;
      static PrintWriter out;
      static final long mod = (long)1e9+7;
      static final long inf = (long)1e18;
      static final int _inf = (int)1e9;
      static final int maxN = (int)1e6;
      static long[] fact, inv;
      
      static class cpr implements Comparator<int[]>{
      	public int compare(int[] a, int[] b) {
      		return Integer.compare(a[0], b[0]);
      	}
      }
      
      static void sort(int[] a){
          List<Integer> A = new ArrayList<>();
          for(int i : a) A.add(i);
          Collections.sort(A);
          for(int i = 0; i < A.size(); i++) a[i] = A.get(i);
      }
      
      static void sort(long[] a){
          List<Long> A = new ArrayList<>();
          for(long i : a) A.add(i);
          Collections.sort(A);
          for(int i = 0; i < A.size(); i++) a[i] = A.get(i);
      }
      
      static void print(int[] a){
          StringBuilder sb = new StringBuilder();
          for(int i = 0; i < a.length; i++) sb.append(a[i] + " ");
          sop(sb);
      }
      
      static void print(long[] a){
          StringBuilder sb = new StringBuilder();
          for(int i = 0; i < a.length; i++) sb.append(a[i] + " ");
          sop(sb);
      }
      
      static int swap(int itself, int dummy){return itself;}
      static long swap(long itself, long dummy){return itself;}
      static char swap(char itself, char dummy){return itself;}
      static void sop(Object o){out.println(o);}
      static int ni(){return f.ni();}
      static long nl(){return f.nl();}
      static double nd(){return f.nd();}
      static String next(){return f.next();}
      static String ns(){return f.ns();}
      static String yes = "YES", no = "NO";
      static char[] nc(){return f.nc();}
      static int[] arr(int len){return f.arr(len);}
      static int gcd(int a, int b){if(b == 0) return a; return gcd(b, a%b);}
      static long gcd(long a, long b){if(b == 0) return a; return gcd(b, a%b);}
      static int fcm(int a, int b){return (a*b)/gcd(a, b);}
      static long fcm(long a, long b){return (a*b)/gcd(a, b);}
      static double log2(long N) {return Math.log(N) / Math.log(2);}
      static void debug(int[] a) {sop(Arrays.toString(a));}
      static void debug(long[] a) {sop(Arrays.toString(a));}
      
      static class Flash
      {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
          
          String next(){
              while(!st.hasMoreTokens()){
                  try{
                      st = new StringTokenizer(br.readLine());
                  }catch(IOException e){
                      e.printStackTrace();
                  }
              }
              
              return st.nextToken();
          }
          
          String ns(){
              String s = new String();
              try{
                  s = br.readLine().trim();
              }catch(IOException e){
                  e.printStackTrace();
              }
              
              return s;
          }
          
          int[] arr(int n){
              int[] a = new int[n];
              for(int i = 0; i < n; i++) a[i] = ni();
              return a;
          }
          
          char[] nc(){return ns().toCharArray();}
          int ni(){return Integer.parseInt(next());}
          long nl(){return Long.parseLong(next());}
          double nd(){return Double.parseDouble(next());}
      }
  }