#include <bits/stdc++.h>
using namespace std;
const bool kami_loc = false;
const string kami_fi = "kamirulez.inp";
const string kami_fo = "kamirulez.out";
void file_io(string fi, string fo) {
  if (fi != "" && (!kami_loc || fi == kami_fi)) {
    freopen(fi.c_str(), "r", stdin);
  }
  if (fo != "" && (!kami_loc || fo == kami_fo)) {
    freopen(fo.c_str(), "w", stdout);
  }
}
void set_up() {
  if (kami_loc) {
    file_io(kami_fi, kami_fo);
  }
  cin.clear();
  cout.clear();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void just_do_it();
void just_exec_it() {
  if (kami_loc) {
    auto pstart = chrono::steady_clock::now();
    just_do_it();
    auto pend = chrono::steady_clock::now();
    int ptime =
        chrono::duration_cast<chrono::milliseconds>(pend - pstart).count();
    string bar(50, '=');
    cout << '\n' << bar << '\n';
    cout << "Time: " << ptime << " ms" << '\n';
  } else {
    just_do_it();
  }
}
int main() {
  set_up();
  just_exec_it();
}
void query() {
  int m, n;
  cin >> m >> n;
  int a[m][n];
  int t = 1e9 + 7;
  int s, f1, f2;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
    }
  }
  for (int j = 0; j < n; j++) {
    s = -1;
    for (int i = 0; i < m; i++) {
      s = max(s, a[i][j]);
    }
    t = min(s, t);
  }
  int res = -1;
  for (int i = 0; i < m; i++) {
    f1 = -1;
    f2 = -1;
    for (int j = 0; j < n; j++) {
      if (a[i][j] > f1) {
        f2 = f1;
        f1 = a[i][j];
      } else if (a[i][j] > f2) {
        f2 = a[i][j];
      }
    }
    if (f2 > t) {
      res = max(res, t);
    } else {
      res = max(res, f2);
    }
  }
  cout << res << '\n';
}
void just_do_it() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    query();
  }
}

### Prompt

Create a solution in Cpp for 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 bool kami_loc = false;
const string kami_fi = "kamirulez.inp";
const string kami_fo = "kamirulez.out";
void file_io(string fi, string fo) {
  if (fi != "" && (!kami_loc || fi == kami_fi)) {
    freopen(fi.c_str(), "r", stdin);
  }
  if (fo != "" && (!kami_loc || fo == kami_fo)) {
    freopen(fo.c_str(), "w", stdout);
  }
}
void set_up() {
  if (kami_loc) {
    file_io(kami_fi, kami_fo);
  }
  cin.clear();
  cout.clear();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void just_do_it();
void just_exec_it() {
  if (kami_loc) {
    auto pstart = chrono::steady_clock::now();
    just_do_it();
    auto pend = chrono::steady_clock::now();
    int ptime =
        chrono::duration_cast<chrono::milliseconds>(pend - pstart).count();
    string bar(50, '=');
    cout << '\n' << bar << '\n';
    cout << "Time: " << ptime << " ms" << '\n';
  } else {
    just_do_it();
  }
}
int main() {
  set_up();
  just_exec_it();
}
void query() {
  int m, n;
  cin >> m >> n;
  int a[m][n];
  int t = 1e9 + 7;
  int s, f1, f2;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
    }
  }
  for (int j = 0; j < n; j++) {
    s = -1;
    for (int i = 0; i < m; i++) {
      s = max(s, a[i][j]);
    }
    t = min(s, t);
  }
  int res = -1;
  for (int i = 0; i < m; i++) {
    f1 = -1;
    f2 = -1;
    for (int j = 0; j < n; j++) {
      if (a[i][j] > f1) {
        f2 = f1;
        f1 = a[i][j];
      } else if (a[i][j] > f2) {
        f2 = a[i][j];
      }
    }
    if (f2 > t) {
      res = max(res, t);
    } else {
      res = max(res, f2);
    }
  }
  cout << res << '\n';
}
void just_do_it() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    query();
  }
}

```