#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("-O3")
#pragma GCC optimize("unroll-loops")
long long gcd(long long n, long long n1) {
  while (n != 0 && n1 != 0) {
    if (n > n1)
      n %= n1;
    else
      n1 %= n;
  }
  return n + n1;
}
long long lcm(long long n, long long n1) {
  long long nod = gcd(n, n1);
  return (n / nod) * (n1 / nod) * nod;
}
long long binpow(long long n, long long m) {
  if (m == 0) return 1;
  if (m % 2 == 1)
    return binpow(n, m - 1) * n;
  else {
    long long b = binpow(n, m / 2);
    return b * b;
  }
}
long long INF = 9999999999999999;
long long mod = 998244353;
void solve() {
  int n, m;
  cin >> m >> n;
  vector<vector<long long>> a(m, vector<long long>(n));
  vector<long long> maxs(n, -INF);
  vector<int> ismx(m, 0);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      maxs[j] = max(maxs[j], a[i][j]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (maxs[i] == a[j][i]) {
        ismx[j]++;
      }
    }
  }
  bool u = false;
  for (int i = 0; i < m; i++)
    if (ismx[i] > 1) u = true;
  if (n > m || u) {
    long long mn = INF;
    for (int i = 0; i < n; i++) mn = min(mn, maxs[i]);
    cout << mn << '\n';
    return;
  }
  long long mx = -INF;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      long long mn = INF;
      for (int i1 = 0; i1 < n; i1++)
        if (i1 != i && i1 != j) mn = min(mn, maxs[i1]);
      long long imx = -INF;
      for (int i1 = 0; i1 < m; i1++) {
        imx = max(imx, min(a[i1][i], a[i1][j]));
      }
      mn = min(mn, imx);
      mx = max(mx, mn);
    }
  }
  cout << mx << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
#pragma GCC optimize("-O3")
#pragma GCC optimize("unroll-loops")
long long gcd(long long n, long long n1) {
  while (n != 0 && n1 != 0) {
    if (n > n1)
      n %= n1;
    else
      n1 %= n;
  }
  return n + n1;
}
long long lcm(long long n, long long n1) {
  long long nod = gcd(n, n1);
  return (n / nod) * (n1 / nod) * nod;
}
long long binpow(long long n, long long m) {
  if (m == 0) return 1;
  if (m % 2 == 1)
    return binpow(n, m - 1) * n;
  else {
    long long b = binpow(n, m / 2);
    return b * b;
  }
}
long long INF = 9999999999999999;
long long mod = 998244353;
void solve() {
  int n, m;
  cin >> m >> n;
  vector<vector<long long>> a(m, vector<long long>(n));
  vector<long long> maxs(n, -INF);
  vector<int> ismx(m, 0);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      maxs[j] = max(maxs[j], a[i][j]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (maxs[i] == a[j][i]) {
        ismx[j]++;
      }
    }
  }
  bool u = false;
  for (int i = 0; i < m; i++)
    if (ismx[i] > 1) u = true;
  if (n > m || u) {
    long long mn = INF;
    for (int i = 0; i < n; i++) mn = min(mn, maxs[i]);
    cout << mn << '\n';
    return;
  }
  long long mx = -INF;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      long long mn = INF;
      for (int i1 = 0; i1 < n; i1++)
        if (i1 != i && i1 != j) mn = min(mn, maxs[i1]);
      long long imx = -INF;
      for (int i1 = 0; i1 < m; i1++) {
        imx = max(imx, min(a[i1][i], a[i1][j]));
      }
      mn = min(mn, imx);
      mx = max(mx, mn);
    }
  }
  cout << mx << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

```