#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.141592653589793238463;
inline bool checkBit(long long n, long long i) {
  return n & ((long long)1 << i);
}
inline long long setBit(long long n, long long i) {
  return n | ((long long)1 << i);
}
inline long long resetBit(long long n, long long i) {
  return n & (~((long long)1 << i));
}
inline void normal(long long& a) { a = (a + 1000000007) % 1000000007; }
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInv(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInv(b));
}
inline long long fact(long long r) {
  long long ans = 1;
  while (r) ans = modMul(ans, r), r--;
  return ans;
}
inline long long invfact(long long r) {
  long long ans = 1;
  while (r) ans = modMul(ans, modInv(r)), r--;
  return ans;
}
inline double dgcd(double a, double b) {
  if (a < b) return dgcd(b, a);
  if (b < .0001) return a;
  return dgcd(b, fmod(a, b));
}
inline long long toint(const std::string& s) {
  std::stringstream ss;
  ss << s;
  long long x;
  ss >> x;
  return x;
}
inline std::string tostring(long long number) {
  std::stringstream ss;
  ss << number;
  return ss.str();
}
inline std::string tobin(long long x) { return std::bitset<63>(x).to_string(); }
clock_t start;
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& q) {
  in >> q.first >> q.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2>& q) {
  out << q.first << " " << q.second;
  return out;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first + p2.first, p1.second + p2.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first - p2.first, p1.second - p2.second};
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T>
T allmin(T t) {
  return t;
}
template <typename T, typename... Args>
T allmin(T t, Args... args) {
  return min(t, allmin(args...));
}
template <typename T>
T allmax(T t) {
  return t;
}
template <typename T, typename... Args>
T allmax(T t, Args... args) {
  return max(t, allmax(args...));
}
template <typename T>
void Unique(vector<T>& v) {
  sort((v).begin(), (v).end()),
      v.resize(distance(v.begin(), unique((v).begin(), (v).end())));
}
const long long N = 2e5;
const long long INF = 2e18;
long long power(long long base, long long exp) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp /= 2;
  }
  return res % 1000000007;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void Timelimitexceeded() {
  long long m, n;
  cin >> m >> n;
  long long p[m][n];
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      cin >> p[i][j];
    }
  }
  long long lo = 0, hi = 1e9;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    long long temp[m][n];
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) {
        temp[i][j] = ((p[i][j] < mid) ? 0 : 1);
      }
    }
    bool flag = true;
    for (long long i = 0; i < n; i++) {
      long long mx = 0;
      for (long long j = 0; j < m; j++) {
        mx |= temp[j][i];
      }
      if (!mx) flag = false;
    }
    bool flag1 = false;
    for (long long i = 0; i < m; i++) {
      long long val = 0;
      for (long long j = 0; j < n; j++) {
        val += temp[i][j];
      }
      if (val > 1) flag1 = true;
    }
    if (flag && flag1) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  cout << lo - 1 << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  clock_t start, end;
  start = clock();
  cout << fixed << setprecision(20);
  cin >> t;
  while (t--) {
    Timelimitexceeded();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}

### Prompt

Develop a solution in cpp 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;
const long double pi = 3.141592653589793238463;
inline bool checkBit(long long n, long long i) {
  return n & ((long long)1 << i);
}
inline long long setBit(long long n, long long i) {
  return n | ((long long)1 << i);
}
inline long long resetBit(long long n, long long i) {
  return n & (~((long long)1 << i));
}
inline void normal(long long& a) { a = (a + 1000000007) % 1000000007; }
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInv(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInv(b));
}
inline long long fact(long long r) {
  long long ans = 1;
  while (r) ans = modMul(ans, r), r--;
  return ans;
}
inline long long invfact(long long r) {
  long long ans = 1;
  while (r) ans = modMul(ans, modInv(r)), r--;
  return ans;
}
inline double dgcd(double a, double b) {
  if (a < b) return dgcd(b, a);
  if (b < .0001) return a;
  return dgcd(b, fmod(a, b));
}
inline long long toint(const std::string& s) {
  std::stringstream ss;
  ss << s;
  long long x;
  ss >> x;
  return x;
}
inline std::string tostring(long long number) {
  std::stringstream ss;
  ss << number;
  return ss.str();
}
inline std::string tobin(long long x) { return std::bitset<63>(x).to_string(); }
clock_t start;
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& q) {
  in >> q.first >> q.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2>& q) {
  out << q.first << " " << q.second;
  return out;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first + p2.first, p1.second + p2.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first - p2.first, p1.second - p2.second};
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T>
T allmin(T t) {
  return t;
}
template <typename T, typename... Args>
T allmin(T t, Args... args) {
  return min(t, allmin(args...));
}
template <typename T>
T allmax(T t) {
  return t;
}
template <typename T, typename... Args>
T allmax(T t, Args... args) {
  return max(t, allmax(args...));
}
template <typename T>
void Unique(vector<T>& v) {
  sort((v).begin(), (v).end()),
      v.resize(distance(v.begin(), unique((v).begin(), (v).end())));
}
const long long N = 2e5;
const long long INF = 2e18;
long long power(long long base, long long exp) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp /= 2;
  }
  return res % 1000000007;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void Timelimitexceeded() {
  long long m, n;
  cin >> m >> n;
  long long p[m][n];
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      cin >> p[i][j];
    }
  }
  long long lo = 0, hi = 1e9;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    long long temp[m][n];
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) {
        temp[i][j] = ((p[i][j] < mid) ? 0 : 1);
      }
    }
    bool flag = true;
    for (long long i = 0; i < n; i++) {
      long long mx = 0;
      for (long long j = 0; j < m; j++) {
        mx |= temp[j][i];
      }
      if (!mx) flag = false;
    }
    bool flag1 = false;
    for (long long i = 0; i < m; i++) {
      long long val = 0;
      for (long long j = 0; j < n; j++) {
        val += temp[i][j];
      }
      if (val > 1) flag1 = true;
    }
    if (flag && flag1) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  cout << lo - 1 << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  clock_t start, end;
  start = clock();
  cout << fixed << setprecision(20);
  cin >> t;
  while (t--) {
    Timelimitexceeded();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}

```