#include <bits/stdc++.h>
using namespace std;
template <class Temp>
vector<Temp> read(long long int n) {
  vector<Temp> input;
  for (long long int i = 0; i < n; i++) {
    Temp k;
    cin >> k;
    input.push_back(k);
  }
  return input;
}
template <class Temp>
vector<vector<Temp>> read_matrix(long long int n, long long int m) {
  vector<vector<Temp>> input(n + 1, vector<Temp>(m + 1, 0));
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++) {
      Temp k;
      cin >> k;
      input[i][j] = k;
    }
  }
  return input;
}
template <class Temp>
void print(vector<long long int> &v) {
  for (Temp a : v) {
    cout << a << " ";
  }
  cout << endl;
}
vector<long int> coor;
bool comparator(long int &i, long int &j) {
  return abs(coor[i]) > abs(coor[j]);
}
void solve() {
  int n;
  cin >> n;
  vector<long long int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  long long int ans = -1;
  for (int i = 1; i < n - 1; i++) {
    if (a[i - 1] <= a[i] && a[i] >= a[i + 1]) {
      long long int t = a[i] * a[i - 1];
      t = max(t, a[i] * a[i + 1]);
      ans = max(ans, t);
    }
  }
  long long int m = max(a[0] * a[1], a[n - 1] * a[n - 2]);
  ans = max(m, ans);
  cout << ans << endl;
}
void test() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  test();
}

### Prompt

Generate a cpp solution to the following problem:

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;
template <class Temp>
vector<Temp> read(long long int n) {
  vector<Temp> input;
  for (long long int i = 0; i < n; i++) {
    Temp k;
    cin >> k;
    input.push_back(k);
  }
  return input;
}
template <class Temp>
vector<vector<Temp>> read_matrix(long long int n, long long int m) {
  vector<vector<Temp>> input(n + 1, vector<Temp>(m + 1, 0));
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++) {
      Temp k;
      cin >> k;
      input[i][j] = k;
    }
  }
  return input;
}
template <class Temp>
void print(vector<long long int> &v) {
  for (Temp a : v) {
    cout << a << " ";
  }
  cout << endl;
}
vector<long int> coor;
bool comparator(long int &i, long int &j) {
  return abs(coor[i]) > abs(coor[j]);
}
void solve() {
  int n;
  cin >> n;
  vector<long long int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  long long int ans = -1;
  for (int i = 1; i < n - 1; i++) {
    if (a[i - 1] <= a[i] && a[i] >= a[i + 1]) {
      long long int t = a[i] * a[i - 1];
      t = max(t, a[i] * a[i + 1]);
      ans = max(ans, t);
    }
  }
  long long int m = max(a[0] * a[1], a[n - 1] * a[n - 2]);
  ans = max(m, ans);
  cout << ans << endl;
}
void test() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  test();
}

```