#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
long long int pow_mod(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long int inv_mod(long long int a) { return pow_mod(a, mod - 2); }
void input(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cin >> a[i];
}
void output(long long int a[], long long int n, bool spaces = true) {
  for (long long int i = 0; i < n; i++) {
    cout << a[i];
    if (!spaces)
      cout << "\n";
    else
      cout << " ";
  }
  if (spaces) cout << endl;
}
void solve() {
  long long int m, n;
  cin >> m >> n;
  long long int p[m][n];
  bool f = false;
  for (long long int i = 0; i < m; i++) {
    for (long long int j = 0; j < n; j++) {
      cin >> p[i][j];
    }
  }
  long long int ans = 0;
  ans = (long long int)1e18;
  for (long long int j = 0; j < n; j++) {
    long long int res = 0;
    for (long long int i = 0; i < m; i++) res = max(res, p[i][j]);
    ans = min(ans, res);
  }
  if (m >= n) {
    long long int mx = 0;
    for (long long int i = 0; i < m; i++) {
      for (long long int j = 0; j < n; j++) {
        for (long long int k = j + 1; k < n; k++)
          mx = max(mx, min(p[i][j], p[i][k]));
      }
    }
    if (mx < ans) ans = mx;
  }
  cout << ans << endl;
}
int main() {
  long long int t;
  t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Your task is to create 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;
const long long int mod = 1000000007;
long long int pow_mod(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long int inv_mod(long long int a) { return pow_mod(a, mod - 2); }
void input(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cin >> a[i];
}
void output(long long int a[], long long int n, bool spaces = true) {
  for (long long int i = 0; i < n; i++) {
    cout << a[i];
    if (!spaces)
      cout << "\n";
    else
      cout << " ";
  }
  if (spaces) cout << endl;
}
void solve() {
  long long int m, n;
  cin >> m >> n;
  long long int p[m][n];
  bool f = false;
  for (long long int i = 0; i < m; i++) {
    for (long long int j = 0; j < n; j++) {
      cin >> p[i][j];
    }
  }
  long long int ans = 0;
  ans = (long long int)1e18;
  for (long long int j = 0; j < n; j++) {
    long long int res = 0;
    for (long long int i = 0; i < m; i++) res = max(res, p[i][j]);
    ans = min(ans, res);
  }
  if (m >= n) {
    long long int mx = 0;
    for (long long int i = 0; i < m; i++) {
      for (long long int j = 0; j < n; j++) {
        for (long long int k = j + 1; k < n; k++)
          mx = max(mx, min(p[i][j], p[i][k]));
      }
    }
    if (mx < ans) ans = mx;
  }
  cout << ans << endl;
}
int main() {
  long long int t;
  t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```