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.*;
  import java.math.*;
  
  public class A4 {
  
      void solve() {
          int m = cint();
          int n = cint();
          int[][] a = new int[m][n];
          int[][] b = new int[m][n];
          int[] c = new int[n];
          for (int i = 0; i < m; i++) {
              for (int j = 0; j < n; j++) {
                  a[i][j] = cint();
                  b[i][j] = a[i][j];
                  c[j] = Math.max(c[j], a[i][j]);
              }
              Arrays.sort(b[i]);
          }
  
          int res = 0;
          for (int i = 0; i < m; i++) {
              res = Math.max(res, b[i][n-2]);
          }
  
          for (int i = 0; i < n; i++) {
              res = Math.min(res, c[i]);
          }
  //        
  //        int[] w = new int[n];
  //        int[] iw = new int[n];
  //        for (int i = 0; i < m; i++) {
  //            for (int j = 0; j < n; j++) {
  //                if (a[i][j] > w[j]) {
  //                    w[j] = a[i][j];
  //                    iw[j] = i;
  //                }
  //            }
  //        }
  //        
  //
  //        int[][] b = new int[n][2];
  //        for (int j = 0; j < n; j++) {
  //            b[j][0] = a[0][j];
  //            b[j][1] = a[1][j];
  //            Arrays.sort(b[j]);
  //            for (int i = 2; i < m; i++) {
  //                if (a[i][j] > b[j][1]) {
  //                    b[j][0] = b[j][1];
  //                    b[j][1] = a[i][j];
  //                } else if (a[i][j] > b[j][0]) {
  //                    b[j][0] = a[i][j];
  //                }
  //            }
  //        }
  //        
  //        int res = 0;
  //        for (int skip = 0; skip < n; skip++) {
  //            int min = Integer.MAX_VALUE;
  //            int max4skip = 0;
  //            for (int i = 0; i < n; i++) {
  //                if (i != skip) {
  //                    min = Math.min(min, w[i]);
  //                    max4skip = Math.max(max4skip, a[iw[i]][skip]);
  //                }
  //            }
  //            min = Math.min(min, max4skip);
  //            res = Math.max(res, min);
  //        }
          cout(res);
      }
  
      public static void main(String... args) {
          int t = in.nextInt();
          for (int test = 0; test < t; test++) {
              new A4().solve();
          }
          out.flush();
      }
  
      static final QuickReader in = new QuickReader(System.in);
      static final PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out, 131_072));
  
      static int cint() {
          return in.nextInt();
      }
  
      static long clong() {
          return in.nextLong();
      }
  
      static String cstr() {
          return in.nextLine();
      }
  
      static int[] carr_int(int n) {
          return in.nextInts(n);
      }
  
      static long[] carr_long(int n) {
          return in.nextLongs(n);
      }
  
      static void cout(int... a) {
          StringBuilder buf = new StringBuilder();
          for (int i = 0; i < a.length; i++) {
              if (i != 0) {
                  buf.append(' ');
              }
              buf.append(a[i]);
          }
          out.println(buf);
      }
  
      static void cout(long... a) {
          StringBuilder buf = new StringBuilder();
          for (int i = 0; i < a.length; i++) {
              if (i != 0) {
                  buf.append(' ');
              }
              buf.append(a[i]);
          }
          out.println(buf);
      }
  
      static void cout(Object o) {
          out.println(o);
      }
  
      static class QuickReader {
          BufferedReader in;
          StringTokenizer token;
  
          public QuickReader(InputStream ins) {
              in = new BufferedReader(new InputStreamReader(ins));
              token = new StringTokenizer("");
          }
  
          public boolean hasNext() {
              while (!token.hasMoreTokens()) {
                  try {
                      String s = in.readLine();
                      if (s == null) {
                          return false;
                      }
                      token = new StringTokenizer(s);
                  } catch (IOException e) {
                      throw new InputMismatchException();
                  }
              }
              return true;
          }
  
          public String next() {
              hasNext();
              return token.nextToken();
          }
  
          public String nextLine() {
              try {
                  String s = in.readLine();
                  token = new StringTokenizer("");
                  return s;
              } catch (IOException e) {
                  throw new InputMismatchException();
              }
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public int[] nextInts(int n) {
              int[] res = new int[n];
              for (int i = 0; i < n; i++) {
                  res[i] = nextInt();
              }
              return res;
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
  
          public long[] nextLongs(int n) {
              long[] res = new long[n];
              for (int i = 0; i < n; i++) {
                  res[i] = nextLong();
              }
              return res;
          }
      }
  }
  