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 sys import setrecursionlimit
  
  from bisect import bisect_left, bisect_right
  from collections import deque
  from functools import lru_cache, reduce
  from heapq import heappush, heappop
  from math import sqrt, ceil, floor, log2
  
  T = int(input())
  
  def rl(t = int):
      return list(map(t, input().split()))
  
  for t in range(1, T + 1):
      rl()
      m, n = rl()
      grid = []
      for _ in range(m):
          grid.append(rl())
  
      # seconds[i] = [j]
      # i is the index of the ith friend
      # j is the index of the shop that
      # has the second highest val for friend i
      seconds = []
      # bests[j] = set(i1, i2, i3, ...)
      # shop j is the best happiness for
      # all friends with indices i[1-3]
      bests = [set() for _ in range(m)]
      ideal = float('inf')
      for i in range(n):
          b1 = b2 = 0
          i1 = i2 = -1
          for j in range(m):
              this = grid[j][i]
              if this > b1:
                  b1, b2 = this, b1
                  i1, i2 = j, i1
              elif this > b2:
                  b2 = this
                  i2 = j
              else:
                  pass
          ideal = min(ideal, b1)
          bests[i1].add(i)
          seconds.append(i2)
  
      if any(len(b) > 1 for b in bests):
          # at least one of these shops has the best
          # choice for >=two people, so we can
          # visit this one and any n-2 other shops we want
          # this is ideal
          print(ideal)
      else:
          # if everyone's ideal item is from a different shop,
          # we have to visit one shop for two people regardless. find the best
          # try to use index cj as the common shop
          ret = 0
          for cj in range(m):
              b1 = b2 = 0
              for el in grid[cj]:
                  if el > b1:
                      b1, b2 = el, b1
                  elif el > b2:
                      b2 = el
                  else:
                      pass
  
              ret = max(ret, min([ideal, b1, b2]))
          print(ret)
  