#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int m, n;
    scanf("%d %d", &m, &n);
    int happyness[m][n];
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        cin >> happyness[i][j];
      }
    }
    if (m == 1) {
      int min = happyness[0][0];
      for (int i = 1; i < n; ++i) {
        if (happyness[0][i] < min) min = happyness[0][i];
      }
      cout << min << endl;
    } else {
      int special_row = 0, ans1 = 0, ans2 = 0, k = 0, column1 = 0, column2 = 1;
      for (int i = 0; i < m; ++i) {
        int max, second_max, max_column, second_max_column;
        if (happyness[i][0] >= happyness[i][1]) {
          max = happyness[i][0], second_max = happyness[i][1];
          max_column = 0, second_max_column = 1;
        } else {
          second_max = happyness[i][0], max = happyness[i][1];
          max_column = 1, second_max_column = 0;
        }
        if (n > 2) {
          for (int j = 2; j < n; ++j) {
            if (happyness[i][j] > max) {
              second_max = max, max = happyness[i][j];
              second_max_column = max_column, max_column = j;
            } else if (happyness[i][j] > second_max && happyness[i][j] <= max) {
              second_max = happyness[i][j];
              second_max_column = j;
            }
          }
        }
        if (second_max > ans1) {
          ans1 = second_max, column1 = second_max_column, column2 = max_column;
        }
      }
      if (n > 2 && m > 1) {
        int columnmax[n - 2];
        for (int i = 0; i < n; ++i) {
          if (i != column1 && i != column2) {
            int j = 0;
            columnmax[k] = happyness[j][i];
            for (j = 1; j < m; ++j) {
              if (happyness[j][i] > columnmax[k]) {
                columnmax[k] = happyness[j][i];
              }
            }
            ++k;
          }
        }
        ans2 = columnmax[0];
        for (int i = 0; i < n - 2; ++i) {
          if (columnmax[i] < ans2) ans2 = columnmax[i];
        }
        if (ans2 <= ans1)
          cout << ans2 << endl;
        else
          cout << ans1 << endl;
      } else
        cout << ans1 << endl;
    }
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int m, n;
    scanf("%d %d", &m, &n);
    int happyness[m][n];
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        cin >> happyness[i][j];
      }
    }
    if (m == 1) {
      int min = happyness[0][0];
      for (int i = 1; i < n; ++i) {
        if (happyness[0][i] < min) min = happyness[0][i];
      }
      cout << min << endl;
    } else {
      int special_row = 0, ans1 = 0, ans2 = 0, k = 0, column1 = 0, column2 = 1;
      for (int i = 0; i < m; ++i) {
        int max, second_max, max_column, second_max_column;
        if (happyness[i][0] >= happyness[i][1]) {
          max = happyness[i][0], second_max = happyness[i][1];
          max_column = 0, second_max_column = 1;
        } else {
          second_max = happyness[i][0], max = happyness[i][1];
          max_column = 1, second_max_column = 0;
        }
        if (n > 2) {
          for (int j = 2; j < n; ++j) {
            if (happyness[i][j] > max) {
              second_max = max, max = happyness[i][j];
              second_max_column = max_column, max_column = j;
            } else if (happyness[i][j] > second_max && happyness[i][j] <= max) {
              second_max = happyness[i][j];
              second_max_column = j;
            }
          }
        }
        if (second_max > ans1) {
          ans1 = second_max, column1 = second_max_column, column2 = max_column;
        }
      }
      if (n > 2 && m > 1) {
        int columnmax[n - 2];
        for (int i = 0; i < n; ++i) {
          if (i != column1 && i != column2) {
            int j = 0;
            columnmax[k] = happyness[j][i];
            for (j = 1; j < m; ++j) {
              if (happyness[j][i] > columnmax[k]) {
                columnmax[k] = happyness[j][i];
              }
            }
            ++k;
          }
        }
        ans2 = columnmax[0];
        for (int i = 0; i < n - 2; ++i) {
          if (columnmax[i] < ans2) ans2 = columnmax[i];
        }
        if (ans2 <= ans1)
          cout << ans2 << endl;
        else
          cout << ans1 << endl;
      } else
        cout << ans1 << endl;
    }
  }
  return 0;
}

```