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 Contest1619D
  {
      static class InputReader {
          BufferedReader reader;
          StringTokenizer tokenizer;
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
          String next() { // reads in the next string
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
          public int nextInt() { // reads in the next int
              return Integer.parseInt(next());
          }
          public long nextLong() { // reads in the next long
              return Long.parseLong(next());
          }
          public double nextDouble() { // reads in the next double
              return Double.parseDouble(next());
              }
          }
      static InputReader r = new InputReader(System.in);
      static PrintWriter pw = new PrintWriter(System.out);
      static long mod = 1000000007; static int n,m; static long[][] p;
      public static void main(String[] args)
      {
      	int t = r.nextInt();
      	while (t > 0)
      	{
      		t--;
      		m = r.nextInt();
      		n = r.nextInt();
      		p = new long[m][n];
      		for (int i = 0; i < m; i ++)
      		{
      			for (int j = 0; j < n; j ++)
      			{
      				p[i][j] = r.nextLong();
      			}
      		}
      		int shops = Math.min(m, n-1);
      		if (m <= n - 1)
      		{
      			long min = Integer.MAX_VALUE;
      			for (int j = 0; j < n; j ++)
      			{
      				long max = 0;
      				for (int i = 0; i < m; i ++)
      				{
      					max = Math.max(max, p[i][j]);
      				}
      				min = Math.min(min, max);
      			}
      			pw.println(min);
      		}
      		else
      		{
      			pw.println(lastTrue(1,2000000000));
      		}
      	}
          pw.close();
      }
      static boolean f(long x)
      {
      	long min = Integer.MAX_VALUE;
  		for (int j = 0; j < n; j ++)
  		{
  			long max = 0;
  			for (int i = 0; i < m; i ++)
  			{
  				max = Math.max(max, p[i][j]);
  			}
  			min = Math.min(min, max);
  		}
  		if (min < x)
  		{
  			return false;
  		}
  		for (int i = 0; i < m; i ++)
  		{
  			long max = 0;
  			int maxI = -1;
  			for (int j = 0; j < n; j ++)
  			{
  				if (p[i][j] > max)
  				{
  					max = p[i][j];
  					maxI = j;
  				}
  			}
  			long max2 = 0;
  			for (int j = 0; j < n; j ++)
  			{
  				if (p[i][j] > max2 && j != maxI)
  				{
  					max2 = p[i][j];
  				}
  			}
  			if (max2 >= x)
  			{
  				return true;
  			}
  		}
  		return false;
      }
      public static int lastTrue(int lo, int hi) {
          lo--;
          while (lo < hi) {
              int mid = lo + (hi - lo + 1) / 2;
              if (f(mid)) {
                  lo = mid;
              } else {
                  hi = mid - 1;
              }
          }
          return lo;
      }
  }