#include <bits/stdc++.h>
using namespace std;
inline void RI(long long &i) { scanf("%lld", &(i)); }
inline void RVI(vector<long long> &v) {
  for (long long i = 0; i < long long((v).size()); ++i) {
    RI(v[i]);
  }
}
inline void RVVI(vector<vector<long long> > &vv) {
  for (long long i = 0; i < long long((vv).size()); ++i) {
    RVI(vv[i]);
  }
}
inline void WI(const long long &i) { printf("%lld\n", i); }
inline void WVI(const vector<long long> &v, char sep = ' ') {
  for (long long i = 0; i < long long((v).size()); ++i) {
    if (i != 0) {
      printf("%c", sep);
    }
    printf("%lld", v[i]);
  }
  printf("\n");
}
inline void WS(const string &s) { printf("%s\n", s.c_str()); }
inline void WB(bool b, const string &yes, const string &no) {
  if (b) {
    WS(yes);
  } else {
    WS(no);
  }
}
inline void YESNO(bool b) { WB(b, "YES", "NO"); }
inline void RS(string &s) {
  static char buf[1000000];
  scanf("%s", buf);
  s = buf;
}
template <typename T>
inline bool IN(T &S, const typename T::key_type &key) {
  return S.find(key) != S.end();
}
int main(int argc, char *argv[]) {
  long long i, j, k;
  long long t, T;
  RI(T);
  for (t = (0); t < (T); ++t) {
    long long n;
    RI(n);
    vector<long long> a(n);
    RVI(a);
    long long mv = *max_element((a).begin(), (a).end());
    long long ans = -1;
    for (i = (0); i < (n - 1); ++i) {
      ans = max(ans, a[i] * a[i + 1]);
    }
    WI(ans);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

You are given n integers a_1, a_2, …, a_n. Find the maximum value of max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r) over all pairs (l, r) of integers for which 1 ≤ l < r ≤ n.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case, print a single integer — the maximum possible value of the product from the statement.

Example

Input


4
3
2 4 3
4
3 2 3 1
2
69 69
6
719313 273225 402638 473783 804745 323328


Output


12
6
4761
381274500335

Note

Let f(l, r) = max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r).

In the first test case, 

  * f(1, 2) = max(a_1, a_2) ⋅ min(a_1, a_2) = max(2, 4) ⋅ min(2, 4) = 4 ⋅ 2 = 8. 
  * f(1, 3) = max(a_1, a_2, a_3) ⋅ min(a_1, a_2, a_3) = max(2, 4, 3) ⋅ min(2, 4, 3) = 4 ⋅ 2 = 8. 
  * f(2, 3) = max(a_2, a_3) ⋅ min(a_2, a_3) = max(4, 3) ⋅ min(4, 3) = 4 ⋅ 3 = 12. 



So the maximum is f(2, 3) = 12.

In the second test case, the maximum is f(1, 2) = f(1, 3) = f(2, 3) = 6.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline void RI(long long &i) { scanf("%lld", &(i)); }
inline void RVI(vector<long long> &v) {
  for (long long i = 0; i < long long((v).size()); ++i) {
    RI(v[i]);
  }
}
inline void RVVI(vector<vector<long long> > &vv) {
  for (long long i = 0; i < long long((vv).size()); ++i) {
    RVI(vv[i]);
  }
}
inline void WI(const long long &i) { printf("%lld\n", i); }
inline void WVI(const vector<long long> &v, char sep = ' ') {
  for (long long i = 0; i < long long((v).size()); ++i) {
    if (i != 0) {
      printf("%c", sep);
    }
    printf("%lld", v[i]);
  }
  printf("\n");
}
inline void WS(const string &s) { printf("%s\n", s.c_str()); }
inline void WB(bool b, const string &yes, const string &no) {
  if (b) {
    WS(yes);
  } else {
    WS(no);
  }
}
inline void YESNO(bool b) { WB(b, "YES", "NO"); }
inline void RS(string &s) {
  static char buf[1000000];
  scanf("%s", buf);
  s = buf;
}
template <typename T>
inline bool IN(T &S, const typename T::key_type &key) {
  return S.find(key) != S.end();
}
int main(int argc, char *argv[]) {
  long long i, j, k;
  long long t, T;
  RI(T);
  for (t = (0); t < (T); ++t) {
    long long n;
    RI(n);
    vector<long long> a(n);
    RVI(a);
    long long mv = *max_element((a).begin(), (a).end());
    long long ans = -1;
    for (i = (0); i < (n - 1); ++i) {
      ans = max(ans, a[i] * a[i + 1]);
    }
    WI(ans);
  }
  return 0;
}

```