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 static java.lang.Math.*;
  import static java.util.Arrays.*;
  
  public class cf1552c {
  
      public static void main(String[] args) throws IOException {
          int t = ri();
          while (t --> 0) {
              int n = rni(), k = ni(), other[] = new int[2 * n];
              long ans = 0;
              fill(other, -1);
              for (int i = 0; i < k; ++i) {
                  int x = rni() - 1, y = ni() - 1;
                  if (x > y) {
                      int __swap = x;
                      x = y;
                      y = __swap;
                  }
                  other[x] = other[y] = x;
                  boolean[] chords = new boolean[2 * n];
                  int cnt = 0;
                  for (int j = x + 1; j < y; ++j) {
                      if (other[j] != -1) {
                          chords[other[j]] ^= true;
                          if (chords[other[j]]) {
                              ++cnt;
                          } else {
                              --cnt;
                          }
                      }
                  }
                  ans += cnt;
              }
              k = n - k;
              while (k > 0) {
                  int x = -1, y = -1, num = 0;
                  for (int i = 0; i < 2 * n; ++i) {
                      if (other[i] == -1) {
                          if (num == 0) {
                              x = i;
                          } else if (num == k) {
                              y = i;
                          }
                          ++num;
                      }
                  }
                  other[x] = other[y] = x;
                  boolean[] chords = new boolean[2 * n];
                  int cnt = 0;
                  for (int j = x + 1; j < y; ++j) {
                      if (j == 2 * n) {
                          j = 0;
                      }
                      if (other[j] != -1) {
                          chords[other[j]] ^= true;
                          if (chords[other[j]]) {
                              ++cnt;
                          } else {
                              --cnt;
                          }
                      }
                  }
                  ans += cnt;
                  --k;
              }
              prln(ans);
          }
          close();
      }
  
      static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
      static StringTokenizer input;
      static Random __r = new Random();
  
      // references
      // IBIG = 1e9 + 7
      // IMAX ~= 2e9
      // LMAX ~= 9e18
  
      // constants
      static final int IBIG = 1000000007;
      static final int IMAX = 2147483647;
      static final long LMAX = 9223372036854775807L;
      // math util
      static int minof(int a, int b, int c) {return min(a, min(b, c));}
      static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
      static long minof(long a, long b, long c) {return min(a, min(b, c));}
      static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
      static int maxof(int a, int b, int c) {return max(a, max(b, c));}
      static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
      static long maxof(long a, long b, long c) {return max(a, max(b, c));}
      static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
      static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
      static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
      static int fli(double d) {return (int) d;}
      static int cei(double d) {return (int) ceil(d);}
      static long fll(double d) {return (long) d;}
      static long cel(double d) {return (long) ceil(d);}
      static int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
      static long gcd(long a, long b) {return b == 0 ? a : gcd(b, a % b);}
      static int[] exgcd(int a, int b) {if (b == 0) return new int[] {1, 0}; int[] y = exgcd(b, a % b); return new int[] {y[1], y[0] - y[1] * (a / b)};}
      static long[] exgcd(long a, long b) {if (b == 0) return new long[] {1, 0}; long[] y = exgcd(b, a % b); return new long[] {y[1], y[0] - y[1] * (a / b)};}
      static int randInt(int min, int max) {return __r.nextInt(max - min + 1) + min;}
      static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
      // array util
      static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void rsort(int[] a) {shuffle(a); sort(a);}
      static void rsort(long[] a) {shuffle(a); sort(a);}
      static void rsort(double[] a) {shuffle(a); sort(a);}
      static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      // input
      static void r() throws IOException {input = new StringTokenizer(rline());}
      static int ri() throws IOException {return Integer.parseInt(rline());}
      static long rl() throws IOException {return Long.parseLong(rline());}
      static double rd() throws IOException {return Double.parseDouble(rline());}
      static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
      static void ria(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni();}
      static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
      static void riam1(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1;}
      static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
      static void rla(long[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nl();}
      static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
      static void rda(double[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nd();}
      static char[] rcha() throws IOException {return rline().toCharArray();}
      static void rcha(char[] a) throws IOException {int n = a.length, i = 0; for (char c : rline().toCharArray()) a[i++] = c;}
      static String rline() throws IOException {return __i.readLine();}
      static String n() {return input.nextToken();}
      static int rni() throws IOException {r(); return ni();}
      static int ni() {return Integer.parseInt(n());}
      static long rnl() throws IOException {r(); return nl();}
      static long nl() {return Long.parseLong(n());}
      static double rnd() throws IOException {r(); return nd();}
      static double nd() {return Double.parseDouble(n());}
      // output
      static void pr(int i) {__o.print(i);}
      static void prln(int i) {__o.println(i);}
      static void pr(long l) {__o.print(l);}
      static void prln(long l) {__o.println(l);}
      static void pr(double d) {__o.print(d);}
      static void prln(double d) {__o.println(d);}
      static void pr(char c) {__o.print(c);}
      static void prln(char c) {__o.println(c);}
      static void pr(char[] s) {__o.print(new String(s));}
      static void prln(char[] s) {__o.println(new String(s));}
      static void pr(String s) {__o.print(s);}
      static void prln(String s) {__o.println(s);}
      static void pr(Object o) {__o.print(o);}
      static void prln(Object o) {__o.println(o);}
      static void prln() {__o.println();}
      static void pryes() {prln("yes");}
      static void pry() {prln("Yes");}
      static void prY() {prln("YES");}
      static void prno() {prln("no");}
      static void prn() {prln("No");}
      static void prN() {prln("NO");}
      static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
      static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
      static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
      static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
      static void h() {prln("hlfd"); flush();}
      static void flush() {__o.flush();}
      static void close() {__o.close();}
  }