description: Vlad has n friends, for each of whom he wants to buy one gift for the New Year.
  
  There are m shops in the city, in each of which he can buy a gift for any of his friends. If the j-th friend (1 ≤ j ≤ n) receives a gift bought in the shop with the number i (1 ≤ i ≤ m), then the friend receives p_{ij} units of joy. The rectangular table p_{ij} is given in the input.
  
  Vlad has time to visit at most n-1 shops (where n is the number of friends). He chooses which shops he will visit and for which friends he will buy gifts in each of them.
  
  Let the j-th friend receive a_j units of joy from Vlad's gift. Let's find the value α=min\\{a_1, a_2, ..., a_n\}. Vlad's goal is to buy gifts so that the value of α is as large as possible. In other words, Vlad wants to maximize the minimum of the joys of his friends.
  
  For example, let m = 2, n = 2. Let the joy from the gifts that we can buy in the first shop: p_{11} = 1, p_{12}=2, in the second shop: p_{21} = 3, p_{22}=4.
  
  Then it is enough for Vlad to go only to the second shop and buy a gift for the first friend, bringing joy 3, and for the second — bringing joy 4. In this case, the value α will be equal to min\{3, 4\} = 3
  
  Help Vlad choose gifts for his friends so that the value of α is as high as possible. Please note that each friend must receive one gift. Vlad can visit at most n-1 shops (where n is the number of friends). In the shop, he can buy any number of gifts.
  
  Input
  
  The first line of the input contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input.
  
  An empty line is written before each test case. Then there is a line containing integers m and n (2 ≤ n, 2 ≤ n ⋅ m ≤ 10^5) separated by a space — the number of shops and the number of friends, where n ⋅ m is the product of n and m.
  
  Then m lines follow, each containing n numbers. The number in the i-th row of the j-th column p_{ij} (1 ≤ p_{ij} ≤ 10^9) is the joy of the product intended for friend number j in shop number i.
  
  It is guaranteed that the sum of the values n ⋅ m over all test cases in the test does not exceed 10^5.
  
  Output
  
  Print t lines, each line must contain the answer to the corresponding test case — the maximum possible value of α, where α is the minimum of the joys from a gift for all of Vlad's friends.
  
  Example
  
  Input
  
  
  5
  
  2 2
  1 2
  3 4
  
  4 3
  1 3 1
  3 1 1
  1 2 2
  1 1 3
  
  2 3
  5 3 4
  2 5 1
  
  4 2
  7 9
  8 1
  9 6
  10 8
  
  2 4
  6 5 2 1
  7 9 7 2
  
  
  Output
  
  
  3
  2
  4
  8
  2
solution: from collections import deque
  import heapq
  import sys
  
  
  def srch(arr, val, idx):
      for id, i in enumerate(arr):
          if id == idx:
              continue
          if i >= val:
              return True
      return False
  
  
  def secondmax(arr):
      mx = max(arr[0], arr[1])
      id_mx = max(range(2), key=arr.__getitem__)
      smx = min(arr[0], arr[1])
      id_smx = min(range(2), key=arr.__getitem__)
      for i in range(2, len(arr)):
          if arr[i] >= mx:
              smx = mx
              id_smx = id_mx
              mx = arr[i]
              id_mx = i
          elif arr[i] >= smx:
              smx = arr[i]
              id_smx = i
      return smx, id_smx, mx, id_mx
  
  
  def solve1(m, n, arr):
      mx_vals = []
      for i in range(m):
          mx_vals.append(secondmax(arr[i]))
      smx, id_smx, mx, id_mx = max(mx_vals)
      vals = [smx, mx]
      carr = list(zip(*arr))
      for i in range(n):
          if i == id_smx or i == id_mx:
              continue
          val = max(carr[i])
          vals.append(val)
      return min(vals)
  
  
  def solve(m, n, arr):
      rnos = set()
      a = sys.maxsize
      cols = list(zip(*arr))
      mat = []
      cmat = []
      best = -1
      # Assign each frnd max joy
      for i in range(n):
          rno = max(range(m), key=cols[i].__getitem__)
          rnos.add(rno)
          # cmat.append(rno)
          if cols[i][rno] < a:
              best = i
              a = cols[i][rno]
          mat.append(arr[rno])
      if len(rnos) < n:
          return a
      # assign next best to rest of the rows
      cmat = list(zip(*mat))
      flag = False
      for i in range(n):
          if i == best:
              continue
          tmp_arr = cmat[i]
          if srch(tmp_arr, a, i):
              return a
      smxs = [secondmax(cmat[i]) for i in range(n)]
      a = max(smxs)
      return a
  
  
  if __name__ == "__main__":
      t = int(input())
      for i in range(t):
          _ = input()
          m, n = tuple(map(int, input().rstrip().split()))
          arr = []
          for _ in range(m):
              arr.append(list(map(int, input().rstrip().split())))
          ans = solve1(m, n, arr)
          print(ans)
  