#include <bits/stdc++.h>
using namespace std;
void __print(long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
template <long long D, typename T>
struct vec : public vector<vec<D - 1, T>> {
  static_assert(D >= 1, "Vector dimension must be greater than zero!");
  template <typename... Args>
  vec(long long n = 0, Args... args)
      : vector<vec<D - 1, T>>(n, vec<D - 1, T>(args...)) {}
};
template <typename T>
struct vec<1, T> : public vector<T> {
  vec(long long n = 0, const T &val = T()) : vector<T>(n, val) {}
};
const long long gf = 1e6 + 9;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long m, n;
    cin >> m >> n;
    long long a[m][n];
    long long mx[m];
    for (long long i = 0; i < m; i++)
      for (long long j = 0; j < n; j++) cin >> a[i][j];
    for (long long i = 0; i < m; i++) {
      long long lol = 0;
      for (long long j = 0; j < n; j++) {
        lol = max(lol, a[i][j]);
      }
      mx[i] = lol;
    }
    long long l = 1, r = (1e9 + 5);
    long long ans = 1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      set<long long> s1[n];
      vector<long long> cnt(m);
      long long flag = 0;
      for (long long i = 0; i < n; i++) {
        long long lol = 0;
        for (long long j = 0; j < m; j++) {
          if (a[j][i] >= mid) cnt[j]++, lol++;
        }
        if (!lol) flag = 1;
      }
      sort(cnt.rbegin(), cnt.rend());
      long long var = 0;
      for (long long i = 0; i < min(n - 1, m); i++) {
        var += cnt[i];
      };
      if (var >= n && !flag) {
        ans = mid;
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### 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;
void __print(long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
template <long long D, typename T>
struct vec : public vector<vec<D - 1, T>> {
  static_assert(D >= 1, "Vector dimension must be greater than zero!");
  template <typename... Args>
  vec(long long n = 0, Args... args)
      : vector<vec<D - 1, T>>(n, vec<D - 1, T>(args...)) {}
};
template <typename T>
struct vec<1, T> : public vector<T> {
  vec(long long n = 0, const T &val = T()) : vector<T>(n, val) {}
};
const long long gf = 1e6 + 9;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long m, n;
    cin >> m >> n;
    long long a[m][n];
    long long mx[m];
    for (long long i = 0; i < m; i++)
      for (long long j = 0; j < n; j++) cin >> a[i][j];
    for (long long i = 0; i < m; i++) {
      long long lol = 0;
      for (long long j = 0; j < n; j++) {
        lol = max(lol, a[i][j]);
      }
      mx[i] = lol;
    }
    long long l = 1, r = (1e9 + 5);
    long long ans = 1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      set<long long> s1[n];
      vector<long long> cnt(m);
      long long flag = 0;
      for (long long i = 0; i < n; i++) {
        long long lol = 0;
        for (long long j = 0; j < m; j++) {
          if (a[j][i] >= mid) cnt[j]++, lol++;
        }
        if (!lol) flag = 1;
      }
      sort(cnt.rbegin(), cnt.rend());
      long long var = 0;
      for (long long i = 0; i < min(n - 1, m); i++) {
        var += cnt[i];
      };
      if (var >= n && !flag) {
        ans = mid;
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```