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.Closeable;
  import java.io.DataInputStream;
  import java.io.FileInputStream;
  import java.io.IOException;
  import java.util.Arrays;
  import java.util.Comparator;
  import java.util.HashSet;
  
  public class D {
      public static void main(String[] args) throws IOException {
          FastScanner scanner = new FastScanner();
          int t = scanner.nextInt();
          while (t-- > 0) {
              int m = scanner.nextInt(), n = scanner.nextInt();
              int[][] mat = new int[m][n];
              int[] colMax = new int[n];
              HashSet<Integer> set = new HashSet();
              for (int i = 0; i < m; i++) {
                  for (int j = 0; j < n; j++) {
                      mat[i][j] = scanner.nextInt();
                      colMax[j] = Math.max(colMax[j], mat[i][j]);
                      set.add(mat[i][j]);
                  }
              }
              int[] values = set.stream().mapToInt($ -> $).sorted().toArray();
              int l = 0, r = values.length - 1;
              while (l != r) {
                  int mid = (l + r) >> 1;
                  mid++;
                  int v = values[mid];
  
                  boolean fail = false;
                  for (int i : colMax) if (i < v) {
                      fail = true;
                      break;
                  }
                  if (!fail) {
                      fail = true;
                      for (int i = 0; i < m; i++) {
                          int count = 0;
                          for (int j = 0; j < n; j++) {
                              if (mat[i][j] >= v && ++count >= 2) break;
                          }
                          if (count >= 2) { fail = false; break; }
                      }
                  }
  
                  if (fail) {
                      r = mid - 1;
                  } else {
                      l = mid;
                  }
              }
  
              System.out.println(values[l]);
          }
      }
  
      /**
       * Credits: https://www.geeksforgeeks.org/fast-io-in-java-in-competitive-programming/
       */
      private static class FastScanner implements Closeable {
          final private int BUFFER_SIZE = 1 << 16;
          private DataInputStream din;
          private byte[] buffer;
          private int bufferPointer, bytesRead;
  
          public FastScanner() {
              din = new DataInputStream(System.in);
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public FastScanner(String file_name) throws IOException {
              din = new DataInputStream(new FileInputStream(file_name));
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public String readLine() throws IOException {
              byte[] buf = new byte[64];
              int cnt = 0, c;
              while ((c = read()) != -1) {
                  if (c == '\n' || c == '\r') {
                      if (cnt != 0) {
                          break;
                      } else {
                          continue;
                      }
                  }
                  buf[cnt++] = (byte) c;
              }
              return new String(buf, 0, cnt);
          }
  
          public String next() throws IOException {
              return next(32);
          }
  
          public String next(int len) throws IOException {
              byte[] buf = new byte[len];
              int ptr = 0;
              byte c = read();
              while (c <= ' ') c = read();
              do {
                  if (ptr == buf.length) {
                      buf = Arrays.copyOf(buf, buf.length << 1);
                  }
                  buf[ptr++] = c;
              } while ((c = read()) > ' ');
              if (buf.length == ptr) return new String(buf);
              return new String(Arrays.copyOf(buf, ptr));
          }
  
          public int nextInt() throws IOException {
              int ret = 0;
              byte c = read();
              while (c <= ' ') c = read();
              boolean neg = (c == '-');
              if (neg) c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg) return -ret;
              return ret;
          }
  
          public long nextLong() throws IOException {
              long ret = 0;
              byte c = read();
              while (c <= ' ') c = read();
              boolean neg = (c == '-');
              if (neg) c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg) return -ret;
              return ret;
          }
  
          public double nextDouble() throws IOException {
              double ret = 0, div = 1;
              byte c = read();
              while (c <= ' ') c = read();
              boolean neg = (c == '-');
              if (neg) c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (c == '.') {
                  while ((c = read()) >= '0' && c <= '9') {
                      ret += (c - '0') / (div *= 10);
                  }
              }
              if (neg) return -ret;
              return ret;
          }
  
          private void fillBuffer() throws IOException {
              bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
              if (bytesRead == -1) buffer[0] = -1;
          }
  
          private byte read() throws IOException {
              if (bufferPointer == bytesRead) fillBuffer();
              return buffer[bufferPointer++];
          }
  
          @Override
          public void close() throws IOException {
              if (din == null) return;
              din.close();
          }
      }
  }
  