#include <bits/stdc++.h>
using namespace std;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = 1000000007;
const char nl = '\n';
const long long MX = 100001;
long long powi(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x) % MOD;
    x = (x * x) % MOD;
    n >>= 1;
  }
  return res;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return 0;
  while (n != 1) {
    if (n % 2 != 0) return 0;
    n = n / 2;
  }
  return 1;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
vector<long long> second[100005];
bool check(long long p, vector<vector<long long>>& matrix, long long n,
           long long m) {
  for (long long i = 0; i < (n); i++) {
    for (long long j = 0; j < (m); j++) {
      if (p <= matrix[j][i]) {
        second[i].push_back(j);
      }
    }
  }
  bool first = 0;
  long long cnt = 0;
  for (long long j = 0; j < (m); j++) {
    for (long long i = 0; i < (n); i++) {
      if (second[i].empty()) {
        cnt = 0;
        break;
      }
      bool f2 = 0;
      for (auto u : second[i]) {
        if (u == j) {
          f2 = 1;
          break;
        }
      }
      if (f2 == 1) {
        cnt++;
      }
    }
    if (cnt > 1) return true;
    cnt = 0;
  }
  return false;
}
void solve() {
  long long m, n;
  cin >> m >> n;
  vector<vector<long long>> matrix(m, vector<long long>(n));
  for (long long i = 0; i < (m); i++) {
    for (long long j = 0; j < (n); j++) cin >> matrix[i][j];
  }
  if (n - 1 >= m) {
    long long mini = 1;
    long long cmini = 1e18;
    for (long long j = 0; j < (n); j++) {
      mini = 1;
      for (long long i = 0; i < (m); i++) {
        mini = max(mini, matrix[i][j]);
      }
      cmini = min(mini, cmini);
    }
    cout << cmini << "\n";
    return;
  }
  long long st = 1;
  long long end = 1e10;
  long long ans = 1;
  while (st <= end) {
    for (long long i = 0; i < (n + 1); i++) second[i].clear();
    long long mid = st + (end - st) / 2;
    if (check(mid, matrix, n, m)) {
      st = mid + 1;
      ans = max(ans, mid);
    } else {
      end = mid - 1;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = 1000000007;
const char nl = '\n';
const long long MX = 100001;
long long powi(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x) % MOD;
    x = (x * x) % MOD;
    n >>= 1;
  }
  return res;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return 0;
  while (n != 1) {
    if (n % 2 != 0) return 0;
    n = n / 2;
  }
  return 1;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
vector<long long> second[100005];
bool check(long long p, vector<vector<long long>>& matrix, long long n,
           long long m) {
  for (long long i = 0; i < (n); i++) {
    for (long long j = 0; j < (m); j++) {
      if (p <= matrix[j][i]) {
        second[i].push_back(j);
      }
    }
  }
  bool first = 0;
  long long cnt = 0;
  for (long long j = 0; j < (m); j++) {
    for (long long i = 0; i < (n); i++) {
      if (second[i].empty()) {
        cnt = 0;
        break;
      }
      bool f2 = 0;
      for (auto u : second[i]) {
        if (u == j) {
          f2 = 1;
          break;
        }
      }
      if (f2 == 1) {
        cnt++;
      }
    }
    if (cnt > 1) return true;
    cnt = 0;
  }
  return false;
}
void solve() {
  long long m, n;
  cin >> m >> n;
  vector<vector<long long>> matrix(m, vector<long long>(n));
  for (long long i = 0; i < (m); i++) {
    for (long long j = 0; j < (n); j++) cin >> matrix[i][j];
  }
  if (n - 1 >= m) {
    long long mini = 1;
    long long cmini = 1e18;
    for (long long j = 0; j < (n); j++) {
      mini = 1;
      for (long long i = 0; i < (m); i++) {
        mini = max(mini, matrix[i][j]);
      }
      cmini = min(mini, cmini);
    }
    cout << cmini << "\n";
    return;
  }
  long long st = 1;
  long long end = 1e10;
  long long ans = 1;
  while (st <= end) {
    for (long long i = 0; i < (n + 1); i++) second[i].clear();
    long long mid = st + (end - st) / 2;
    if (check(mid, matrix, n, m)) {
      st = mid + 1;
      ans = max(ans, mid);
    } else {
      end = mid - 1;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```