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: //package newyearsproblem;
  
  import java.util.*;
  import java.io.*;
  
  public class newyearsproblem {
  	
  	public static boolean isValid(int val, int[][] shops, int n, int m) {
  		int[][] nums = new int[m][n];
  		for(int i = 0; i < n; i++) {
  			int count = 0;
  			for(int j = 0; j < m; j++) {
  				if(shops[j][i] >= val) {
  					count ++;
  					nums[j][i] = 1;
  				}
  			}
  			if(count == 0) {	//a person can't be satisfied
  				return false;
  			}
  		}
  		for(int i = 0; i < m; i++) {
  			int count = 0;
  			for(int j = 0; j < n; j++) {
  				if(nums[i][j] == 1) {
  					count ++;
  				}
  			}
  			if(count > 1) {	//there is a shop that can satisfy two or more people
  				return true;
  			}
  		}
  		return false;
  	}
  	
  	public static void main(String[] args) throws IOException {
  		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(fin.readLine());
  		StringBuilder fout = new StringBuilder();
  		while(t-- > 0) {
  			fin.readLine();
  			StringTokenizer st = new StringTokenizer(fin.readLine());
  			int m = Integer.parseInt(st.nextToken());
  			int n = Integer.parseInt(st.nextToken());
  			int[][] shops = new int[m][n];
  			for(int i = 0; i < m; i++) {
  				st = new StringTokenizer(fin.readLine());
  				for(int j = 0; j < n; j++) {
  					shops[i][j] = Integer.parseInt(st.nextToken());
  				}
  			}
  			int ans = 0;
  			int low = 0;
  			int high = 1000000000;
  			int mid = low + (high - low) / 2;
  			while(low <= high) {
  				if(isValid(mid, shops, n, m)) {
  					ans = Math.max(ans, mid);
  					low = mid + 1;
  				}
  				else {
  					high = mid - 1;
  				}
  				mid = low + (high - low) / 2;
  			}
  			fout.append(ans).append("\n");
  		}
  		System.out.print(fout);
  	}
  }
  