#include <bits/stdc++.h>
using namespace std;
long long int i, j, mod = 1e9 + 7;
long long int power(long long int x, long long int y, long long int mod) {
  x = x % mod;
  long long int res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res % mod;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int ncr(long long int n, long long int r, long long int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  unsigned long long fac[n + 1];
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int fact(long long int n) {
  if (n == 0) return 1;
  return n * fact(n - 1);
}
long long int nCr(long long int n, long long int r) {
  if (r == 2)
    return (n * (n - 1)) / 2;
  else
    return (n * (n - 1) * (n - 2)) / 6;
}
bool isPos(vector<vector<long long int>> &arr1, long long int mid) {
  long long int m = arr1.size(), n = arr1[0].size();
  long long int total = 0, extra = 0;
  set<long long int> sel;
  for (i = 0; i < n; i++) {
    total = 0;
    for (j = 0; j < m; j++) {
      if (arr1[j][i] >= mid) {
        total = 1;
        if (sel.count(j)) extra = 1;
        sel.insert(j);
      }
    }
    if (!total) break;
  }
  return total && extra;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long int t;
  cin >> t;
  while (t--) {
    long long int m, n;
    cin >> m >> n;
    vector<vector<long long int>> arr(m, vector<long long int>(n));
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) cin >> arr[i][j];
    }
    long long int strt = 1, end = 1e9, ans = -1;
    while (strt <= end) {
      long long int mid = strt + (end - strt) / 2;
      if (isPos(arr, mid)) {
        ans = mid;
        strt = mid + 1;
      } else
        end = mid - 1;
    }
    cout << ans << endl;
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long int i, j, mod = 1e9 + 7;
long long int power(long long int x, long long int y, long long int mod) {
  x = x % mod;
  long long int res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res % mod;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int ncr(long long int n, long long int r, long long int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  unsigned long long fac[n + 1];
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int fact(long long int n) {
  if (n == 0) return 1;
  return n * fact(n - 1);
}
long long int nCr(long long int n, long long int r) {
  if (r == 2)
    return (n * (n - 1)) / 2;
  else
    return (n * (n - 1) * (n - 2)) / 6;
}
bool isPos(vector<vector<long long int>> &arr1, long long int mid) {
  long long int m = arr1.size(), n = arr1[0].size();
  long long int total = 0, extra = 0;
  set<long long int> sel;
  for (i = 0; i < n; i++) {
    total = 0;
    for (j = 0; j < m; j++) {
      if (arr1[j][i] >= mid) {
        total = 1;
        if (sel.count(j)) extra = 1;
        sel.insert(j);
      }
    }
    if (!total) break;
  }
  return total && extra;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long int t;
  cin >> t;
  while (t--) {
    long long int m, n;
    cin >> m >> n;
    vector<vector<long long int>> arr(m, vector<long long int>(n));
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) cin >> arr[i][j];
    }
    long long int strt = 1, end = 1e9, ans = -1;
    while (strt <= end) {
      long long int mid = strt + (end - strt) / 2;
      if (isPos(arr, mid)) {
        ans = mid;
        strt = mid + 1;
      } else
        end = mid - 1;
    }
    cout << ans << endl;
  }
}

```