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.util.*;
  import java.util.Scanner;
  
  
  public class Solution {
  	static int mod=1000000007;;
  	
  //	 
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  //		System.out.println();
  
  		Scanner sc=new Scanner(System.in);
  		int tt=sc.nextInt();
  		
  		
  		while(tt-->0){
  			
  //
  //		
  			int n=sc.nextInt();
  			int m=sc.nextInt();
  //			long n=sc.nextLong();
  			int arr[][]=new int[n][m];
  			int max[]=new int[m];
  			Arrays.fill(max, 0);
  			for(int i=0;i<n;i++) {
  				for(int j=0;j<m;j++) {
  					arr[i][j]=sc.nextInt();
  					max[j]=Math.max(max[j], arr[i][j]);
  				}
  			}
  			
  //			for(int i=0;i<n;i++) {
  //				for(int j=0;j<m;j++) {
  //					System.out.print(max[j]+" ");
  //				}
  //				System.out.println();
  //			}
  			
  			int ans=0;
  			for(int i=0;i<n;i++) {
  				int maxOne=0;
  				int maxtwo=0;
  				for(int j=0;j<m;j++) {
  					if(arr[i][j]>maxOne) {
  						maxtwo=maxOne;
  						maxOne=arr[i][j];
  					}
  					else if(arr[i][j]>maxtwo) {
  						maxtwo=arr[i][j];
  					}
  						
  				}
  				
  				int min=maxtwo;
  				for(int j=0;j<m;j++) {
  					if(arr[i][j]!=maxOne&&arr[i][j]!=maxtwo) {
  						min=Math.min(max[j], min);
  					}
  				}
  				if(min<maxtwo) {
  					ans=Math.max(ans, min);
  				}
  				else {
  					if(maxOne==maxtwo) {
  						PriorityQueue<Integer>pq=new PriorityQueue<>();
  						for(int j=0;j<m;j++) {
  							if(arr[i][j]==maxOne) {
  								if(max[j]<min) {
  									pq.add(max[j]);
  								}
  								
  							}
  						}
  						
  						pq.poll();
  						pq.poll();
  						if(!pq.isEmpty()) {
  							min=Math.min(pq.poll(), min);
  						}
  					}
  					else {
  						PriorityQueue<Integer>pq=new PriorityQueue<>();
  						for(int j=0;j<m;j++) {
  							if(arr[i][j]==maxtwo) {
  								if(max[j]<min) {
  									pq.add(max[j]);
  								}
  								
  							}
  						}
  						
  						pq.poll();
  						
  						if(!pq.isEmpty()) {
  							min=Math.min(pq.poll(), min);
  						}
  						
  					}
  					ans=Math.max(ans, min);
  				}
  				
  			}
  			
  			
  			
  				
  				System.out.println(ans);
  			
  		
  			
  		
  			
  		}
  	}
  		
  		
  		
  		
  		
  		
  	
  	
  	
  
  }