#include <bits/stdc++.h>
using namespace std;
double PI = (acos(-1));
long long md = 1000000007;
long long pw(long long a, long long b) {
  long long c = 1, m = a;
  while (b) {
    if (b & 1) c = (c * m);
    m = (m * m);
    b /= 2;
  }
  return c;
}
long long pwmd(long long a, long long b) {
  long long c = 1, m = a;
  while (b) {
    if (b & 1) c = ((c * m)) % md;
    m = (m * m) % md;
    b /= 2;
  }
  return c;
}
long long modinv(long long n) { return pwmd(n, md - 2); }
long long nc2(long long n) { return (1ll * n * (n - 1)) / 2; }
bool prime(long long a) {
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
long long ceel(long long a, long long b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
bool cch(long long a, long long b) {
  long double x = a, y = b;
  long double z = 1000000000000000000 / a;
  if (y > z) return 0;
  return 1;
}
void solve() {
  long long m, n;
  cin >> m >> n;
  long long l[m][n];
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) cin >> l[i][j];
  }
  if (n - 1 >= m) {
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      long long mx = 0;
      for (long long j = 0; j < m; j++) {
        mx = max(mx, l[j][i]);
      }
      ans.push_back(mx);
    }
    sort(ans.begin(), ans.end());
    cout << ans[0] << "\n";
    return;
  } else {
    long long x = 1, y = 1000000001;
    while (x + 1 < y) {
      long long mid = (x + y) / 2, flg = 0, gg = 0;
      map<long long, long long> mp;
      for (long long i = 0; i < n; i++) {
        long long zq = 0;
        for (long long j = 0; j < m; j++) {
          if (l[j][i] >= mid) {
            zq = 1;
            if (mp[j]) flg = 1;
            mp[j] = 1;
          }
        }
        if (zq == 0) gg = 1;
      }
      if (flg == 0 || gg) {
        y = mid;
      } else
        x = mid;
    }
    cout << x << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
double PI = (acos(-1));
long long md = 1000000007;
long long pw(long long a, long long b) {
  long long c = 1, m = a;
  while (b) {
    if (b & 1) c = (c * m);
    m = (m * m);
    b /= 2;
  }
  return c;
}
long long pwmd(long long a, long long b) {
  long long c = 1, m = a;
  while (b) {
    if (b & 1) c = ((c * m)) % md;
    m = (m * m) % md;
    b /= 2;
  }
  return c;
}
long long modinv(long long n) { return pwmd(n, md - 2); }
long long nc2(long long n) { return (1ll * n * (n - 1)) / 2; }
bool prime(long long a) {
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
long long ceel(long long a, long long b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
bool cch(long long a, long long b) {
  long double x = a, y = b;
  long double z = 1000000000000000000 / a;
  if (y > z) return 0;
  return 1;
}
void solve() {
  long long m, n;
  cin >> m >> n;
  long long l[m][n];
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) cin >> l[i][j];
  }
  if (n - 1 >= m) {
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      long long mx = 0;
      for (long long j = 0; j < m; j++) {
        mx = max(mx, l[j][i]);
      }
      ans.push_back(mx);
    }
    sort(ans.begin(), ans.end());
    cout << ans[0] << "\n";
    return;
  } else {
    long long x = 1, y = 1000000001;
    while (x + 1 < y) {
      long long mid = (x + y) / 2, flg = 0, gg = 0;
      map<long long, long long> mp;
      for (long long i = 0; i < n; i++) {
        long long zq = 0;
        for (long long j = 0; j < m; j++) {
          if (l[j][i] >= mid) {
            zq = 1;
            if (mp[j]) flg = 1;
            mp[j] = 1;
          }
        }
        if (zq == 0) gg = 1;
      }
      if (flg == 0 || gg) {
        y = mid;
      } else
        x = mid;
    }
    cout << x << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
  }
  return 0;
}

```