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 __future__ import division, print_function
  import math
  import sys
  import os
  from io import BytesIO, IOBase
  from collections import deque, Counter, OrderedDict, defaultdict
  #import heapq
  #ceil,floor,log,sqrt,factorial,pow,pi,gcd
  #import bisect
  #from bisect import bisect_left,bisect_right
  
  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")
  
  
  def print(*args, **kwargs):
  	"""Prints the values to a stream, or to sys.stdout by default."""
  	sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
  	at_start = True
  	for x in args:
  		if not at_start:
  			file.write(sep)
  		file.write(str(x))
  		at_start = False
  	file.write(kwargs.pop("end", "\n"))
  	if kwargs.pop("flush", False):
  		file.flush()
  
  
  if sys.version_info[0] < 3:
  	sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
  	sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  def inp():
      return(int(input()))
  def inps():
      return input().strip()
  def inlt():
      return(list(map(int,input().split())))
  def insr():
      s = input().strip()
      return(list(s[:len(s)]))
  def invr():
      return(map(int,input().split()))
  def check(k):
      # print(k)
      val=[0 for i in range(m)]
      cnt=[0 for i in range(n)]
      for i in range(n):
          for j in range(m):
              if l[i][j]>=k:
                  cnt[i]+=1
                  val[j]=1
      if max(cnt)>1 and sum(val)==m:
          return 1
      return 0
  
  
  for _ in range(inp()):
      inlt()
      n,m=invr()
      l=[]
      for i in range(n):
          l.append(inlt())
      lf,rt=1,10**9
      while(lf<rt-1):
          mid=(lf+rt)//2
          if check(mid):
              lf=mid
          else:
              rt=mid
      if check(rt):
          print(rt)
      else:
          print(lf)
      