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: import java.io.*;
  import java.util.*;
  public class newyear {
  
  	public static void main(String[] args)throws IOException {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int T = Integer.parseInt(br.readLine());
  		while(T-- > 0) {
  			br.readLine();
  			StringTokenizer st = new StringTokenizer(br.readLine());
  			int M = Integer.parseInt(st.nextToken());
  			int N = Integer.parseInt(st.nextToken());
  			int[][] shop = new int[M][N];
  			for(int i = 0;i < M;i++) {
  				st = new StringTokenizer(br.readLine());
  				for(int j = 0;j < N;j++) {
  					shop[i][j] = Integer.parseInt(st.nextToken());
  				}
  			}
  			
  			int lo = 0;
  			int hi = (int) 1e9;
  			while(lo < hi - 4) {
  				int mid = (lo + hi) / 2;
  				if(val(mid, shop, M)) {
  					lo = mid;
  				}
  				else {
  					hi = mid - 1;
  				}
  			}
  			
  			int ans = 0;
  			for(int i = lo;i <= hi;i++) {
  				if(val(i, shop, M)) {
  					ans = i;
  				}
  				else break;
  			}
  			
  			System.out.println(ans);
  		}
  
  	}
  
  	private static boolean val(int mid, int[][] shop, int M) {
  		boolean[] hasGift = new boolean[shop[0].length];
  		boolean at_least_two = false;
  		for(int i = 0;i < shop.length;i++) {
  			int giftCount = 0;
  			for(int j = 0;j < shop[i].length;j++) {
  				if(shop[i][j] >= mid) {
  					hasGift[j] = true;
  					giftCount++;
  				}
  			}
  			if(giftCount > 1) at_least_two = true;
  		}
  		
  		if(!at_least_two) return false;
  		boolean flag = true;
  		for(boolean x: hasGift) {
  			flag &= x;
  		}
  		return flag;
  	}
  
  }
  