#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1000000001;
const long long inf = 1e18 + 1;
const long long MOD = 1000000007;
template <class T>
void __print(T v);
template <class T>
void __print(vector<T> v);
template <class T>
void __print(set<T> v);
template <class T1, class T2>
void __print(map<T1, T2> mp);
template <class T1, class T2>
void __print(pair<T1, T2> p);
template <class T1, class T2>
void __print(unordered_map<T1, T2> mp);
template <class T>
void __print(T v) {
  cerr << v;
}
template <class T>
void __print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    __print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void __print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    __print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T1, class T2>
void __print(map<T1, T2> mp) {
  cerr << "[ ";
  for (auto i : mp) {
    __print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T1, class T2>
void __print(pair<T1, T2> p) {
  cerr << "{ ";
  __print(p.first);
  cerr << " ";
  __print(p.second);
  cerr << " }";
}
template <class T1, class T2>
void __print(unordered_map<T1, T2> mp) {
  cerr << "[ ";
  for (auto i : mp) {
    __print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long binpow(long long a, long long b) {
  a %= MOD;
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
void pranav() {
  long long m, n;
  cin >> m >> n;
  long long v[m][n];
  long long ans = -1, shop = -1;
  pair<long long, long long> ind;
  for (long long i = 0; i < m; i++) {
    vector<pair<long long, long long>> v1;
    for (long long j = 0; j < n; j++) {
      cin >> v[i][j];
      v1.push_back({v[i][j], j});
    }
    sort(v1.begin(), v1.end());
    reverse(v1.begin(), v1.end());
    if (ans < min(v1[0].first, v1[1].first)) {
      shop = i;
      ind.first = v1[0].second;
      ind.second = v1[1].second;
      ans = min(v1[0].first, v1[1].first);
    }
  }
  for (long long i = 0; i < n; i++) {
    long long val = -1;
    if (i == ind.first || i == ind.second) continue;
    for (long long j = 0; j < m; j++) val = max(val, v[j][i]);
    ans = min(ans, val);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    pranav();
  }
  return 0;
}

### Prompt

Generate 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;
const long long MAXN = 1000000001;
const long long inf = 1e18 + 1;
const long long MOD = 1000000007;
template <class T>
void __print(T v);
template <class T>
void __print(vector<T> v);
template <class T>
void __print(set<T> v);
template <class T1, class T2>
void __print(map<T1, T2> mp);
template <class T1, class T2>
void __print(pair<T1, T2> p);
template <class T1, class T2>
void __print(unordered_map<T1, T2> mp);
template <class T>
void __print(T v) {
  cerr << v;
}
template <class T>
void __print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    __print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void __print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    __print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T1, class T2>
void __print(map<T1, T2> mp) {
  cerr << "[ ";
  for (auto i : mp) {
    __print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T1, class T2>
void __print(pair<T1, T2> p) {
  cerr << "{ ";
  __print(p.first);
  cerr << " ";
  __print(p.second);
  cerr << " }";
}
template <class T1, class T2>
void __print(unordered_map<T1, T2> mp) {
  cerr << "[ ";
  for (auto i : mp) {
    __print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long binpow(long long a, long long b) {
  a %= MOD;
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
void pranav() {
  long long m, n;
  cin >> m >> n;
  long long v[m][n];
  long long ans = -1, shop = -1;
  pair<long long, long long> ind;
  for (long long i = 0; i < m; i++) {
    vector<pair<long long, long long>> v1;
    for (long long j = 0; j < n; j++) {
      cin >> v[i][j];
      v1.push_back({v[i][j], j});
    }
    sort(v1.begin(), v1.end());
    reverse(v1.begin(), v1.end());
    if (ans < min(v1[0].first, v1[1].first)) {
      shop = i;
      ind.first = v1[0].second;
      ind.second = v1[1].second;
      ans = min(v1[0].first, v1[1].first);
    }
  }
  for (long long i = 0; i < n; i++) {
    long long val = -1;
    if (i == ind.first || i == ind.second) continue;
    for (long long j = 0; j < m; j++) val = max(val, v[j][i]);
    ans = min(ans, val);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    pranav();
  }
  return 0;
}

```