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.io.*;
  
  public class D {
  
  	static FastScanner fs = new FastScanner();
  	static Scanner scn = new Scanner(System.in);
  	static PrintWriter pw = new PrintWriter(System.out);
  	static StringBuilder sb = new StringBuilder("");
  	static int[][] p;
  	static boolean used[];
  	static int n, m;
  
  	public static void main(String[] args) {
  		int t = fs.nextInt();
  		for (int tt = 0; tt < t; tt++) {
  			m = fs.nextInt();
  			n = fs.nextInt();
  			p = new int[m][n];
  			for (int i = 0; i < m; i++)
  				for (int j = 0; j < n; j++)
  					p[i][j] = fs.nextInt();
  			long l = 0, r = (long) 1e10;
  			while (l < r) {
  				long mid = (l+r)/2;
  				int[] col = new int[n];
  				int found = 0;
  				for (int i = 0; i < m; i++) {
  					int count = 0;
  					for (int j = 0; j < n; j++) {
  						if (p[i][j] >= mid) {
  							col[j] = 1;
  							count++;
  						}
  					}
  					if (count >= 2) found = 1;
  				}
  				int sum = 0;
  				for (int i = 0; i < n; i++) sum += col[i];
  				if (sum == n && found == 1) l = mid+1;
  				else r = mid;
  			}
  			sb.append((l-1) + "\n");
  		}
  		pw.print(sb.toString());
  		pw.close();
  	}
  	
  	static boolean ok(long x) {
  		for (int i = 0; i < n; i++) {
  			for (int j = 0; j < m; j++) {
  				if (p[i][j] >= x) {
  					if (used[j]) return true;
  					used[j] = true;
  				}
  			}
  		}
  		return false;
  	}
  
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  				}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		int[] readArray(int n) {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) {
  				a[i] = nextInt();
  			}
  			return a;
  		}
  	}
  }