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 sys, os
  from io import BytesIO, IOBase
  from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
  from collections import defaultdict as dd, deque
  from heapq import merge, heapify, heappop, heappush, nsmallest
  from bisect import bisect_left as bl, bisect_right as br, bisect
  
  # region fastio
  BUFSIZE = 8192
   
  class FastIO(IOBase):
      newlines = 0
   
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
   
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
   
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
   
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
   
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  stdin, stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) 
  
  mod = pow(10, 9) + 7
  mod2 = 998244353
          
  def inp(): return stdin.readline().strip()
  def iinp(): return int(inp())
  def out(var, end="\n"): stdout.write(str(var)+"\n")
  def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
  def lmp(): return list(mp())
  def mp(): return map(int, inp().split())
  def l1d(n, val=0): return [val for i in range(n)]
  def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
  def ceil(a, b): return (a+b-1)//b
  S1 = 'abcdefghijklmnopqrstuvwxyz'
  S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
  def isprime(x):
      if x<=1: return False
      if x in (2, 3): return True
      if x%2 == 0: return False
      for i in range(3, int(sqrt(x))+1, 2):
          if x%i == 0: return False
      return True
   
  for _ in range(iinp()):
      n, k = mp()
      lines = []
      taken = l1d(2*n)
      for i in range(k):
          a, b = sorted(mp())
          lines.append((a, b))
          taken[a-1] = 1
          taken[b-1] = 1
      new = taken.count(0)//2
      ans = new*(new-1)//2
      acm = 0
      for a, b in lines:
          x = taken[a:b].count(0)
          y = 2*new - x
          ans += min(x, y)
      for i in range(len(lines)):
          a, b = lines[i]
          for j in range(i+1, len(lines)):
              c, d = lines[j]
              if a < c < b < d or c < a < d < b or c < a < d < b or a < c < b < d:
                  acm += 1
      print(ans + acm)