human
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...
 ... Write me a Python program to find the value α=min\\{a_1, a_2,?
assistant
#Here is Python program to find the value α=min\\{a_1, a_2, .

twoPowers = []; bb = 1
while bb < 1e9:
    twoPowers.append(bb)
    bb *= 2
twoPowers.reverse()

def main():
    
    def check(alpha):
        # first check that for every j friend, there's a value >= alpha
        hasLarger = [False] * n
        for i in range(m):
            for j in range(n):
                if p[i][j] >= alpha:
                    hasLarger[j] = True
        if False in hasLarger:
            return False
        # if m >= n, need to check that at least 1 shop has 2 friends >= alpha
        # this is due to pigeonhole principle
        for i in range(m):
            cnts = 0
            for j in range(n):
                if p[i][j] >= alpha:
                    cnts += 1
            if cnts >= 2:
                return True
        return False
    
    t = int(input())
    allans = []
    for _ in range(t):
        input()
        m, n = readIntArr()
        p = [] # p[i][j] is the joy for jth friend from the ith shop
        for __ in range(m):
            p.append(readIntArr())
        ans = 0
        for bb in twoPowers:
            if check(ans + bb):
                ans += bb
        allans.append(ans)
    multiLineArrayPrint(allans)
    
    return

import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.

def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]
 
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
    dv=defaultValFactory;da=dimensionArr
    if len(da)==1:return [dv() for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(a, b, c):
   ...