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 Round762Div3;
  
  import java.util.*;
  
  public class NewYearsProblem {
  	static int m;
  	static int n;
  	static int[][] data;
  
  	// binary search for possible minimum values, check if they can actually be
  	// achieved
  	// the search will end at the largest possible minimum value, which is our
  	// desired value
  	public static int binarySearch(int low, int high) {
  //		System.out.println(low + " " + high);
  		if (low >= high) {
  			return low;
  		}
  		if (low + 1 == high) {
  			if (check(high)) {
  				return high;
  			} else {
  				return low;
  			}
  		}
  //		System.out.println(low + " " + high); 
  		int mid = (low + high) / 2;
  		if (check(mid)) {
  			return binarySearch(mid, high);
  		} else {
  			return binarySearch(low, mid - 1);
  		}
  	}
  
  	public static boolean check(int min) { // checks if minimum joy can be achieved
  		ArrayList<Set<Integer>> pos = new ArrayList<Set<Integer>>();
  		// for each friend, find which shops we need to visit to satisfy his minimum joy
  		for (int i = 0; i < n; i++) {
  			Set<Integer> temp = new HashSet<Integer>();
  			pos.add(temp);
  		}
  
  		for (int i = 0; i < n; i++) {
  			for (int j = 0; j < m; j++) {
  				if (data[j][i] >= min) {
  					pos.get(i).add(j); // add that friend i can possibly get gift from shop j
  				}
  			}
  		}
  //		System.out.println(pos);
  		// traverse through all possible positions
  		for (int i = 0; i < n; i++) {
  			if (pos.get(i).size() == 0) {
  				return false; // if there's a friend with no possible shops
  			}
  		}
  		for (int i = 0; i < n - 1; i++) {
  			for (int j = i + 1; j < n; j++) {
  				Set<Integer> intersection = new HashSet<Integer>(pos.get(i));
  				intersection.retainAll(pos.get(j));
  				if (intersection.size() > 0) {
  					return true;
  				}
  			}
  		}
  		return false;
  	}
  
  	public static void main(String[] args) {
  		Scanner sc = new Scanner(System.in);
  		int t = sc.nextInt();
  
  		int maxJoy = Integer.MIN_VALUE; // stores the upper bound for binary search later
  
  		for (int x = 0; x < t; x++) {
  			m = sc.nextInt(); // # of shops
  			n = sc.nextInt(); // # of friends
  
  			data = new int[m][n];
  			for (int i = 0; i < m; i++) {
  				int[] temp = new int[n];
  				for (int j = 0; j < n; j++) {
  					temp[j] = sc.nextInt();
  					maxJoy = Math.max(maxJoy, temp[j]);
  				}
  				data[i] = temp;
  			}
  
  			System.out.println(binarySearch(0, maxJoy));
  //			System.out.println();
  		}
  
  	}
  
  }
  