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.BufferedReader;
  import java.io.BufferedWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.OutputStreamWriter;
  import java.nio.charset.StandardCharsets;
  import java.util.BitSet;
  
  public class MainD {
      public static void main(String[] args) throws IOException {
          BufferedReader reader = new BufferedReader(new InputStreamReader(System.in, StandardCharsets.UTF_8));
          BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out, StandardCharsets.UTF_8));
          int t = Integer.parseInt(reader.readLine());
          for (int i = 0; i < t; i++) {
              reader.readLine();
              String[] line1 = reader.readLine().split(" ");
              // m 家商店
              int m = Integer.parseInt(line1[0]);
              // n 个好友
              int n = Integer.parseInt(line1[1]);
              int[][] nums = new int[m][n];
              for (int j = 0; j < m; j++) {
                  String[] lineJ = reader.readLine().split(" ");
                  for (int k = 0; k < n; k++) {
                      nums[j][k] = Integer.parseInt(lineJ[k]);
                  }
              }
              writer.write(solution(m, n, nums).concat(System.lineSeparator()));
          }
          writer.close();
          reader.close();
      }
  
      private static String solution(int m, int n, int[][] nums) {
          // 二分
          int min = Integer.MAX_VALUE;
          int max = Integer.MIN_VALUE;
          for (int[] num : nums) {
              for (int i : num) {
                  min = Math.min(min, i);
                  max = Math.max(max, i);
              }
          }
          int left = min;
          int right = max + 1;
          while (left < right) {
              int mid = left + (right - left) / 2;
              // 边界二分 F, F,..., F, [T, T,..., T] checkMid(mid) == T
              // ----------------------^
              if (!checkMid(m, n, nums, mid)) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          // 右边界二分
          return String.valueOf(left - 1);
      }
  
      private static boolean checkMid(int m, int n, int[][] nums, int mid) {
          // 访问最多 n−1 家商店，必然至少 2 个朋友在同 1 家商店购物
          boolean gt1 = false;
          BitSet bitSet = new BitSet(n);
          for (int i = 0; i < m; i++) {
              int cnt = 0;
              for (int j = 0; j < n; j++) {
                  // nums[i][j] 表示在商店 i 购买 j 朋友的快乐值
                  if (nums[i][j] >= mid) {
                      bitSet.set(j);
                      cnt++;
                      if (cnt > 1) {
                          gt1 = true;
                      }
                  }
              }
          }
          return gt1 && bitSet.cardinality() == n;
      }
  }