#include <bits/stdc++.h>
using namespace std;
long long int ksm(long long int a, long long int b, long long int mod) {
  long long int ans = 1;
  while (b) {
    if (b & 1) {
      ans = (a % mod * ans % mod) % mod;
    }
    b >>= 1;
    a = (a % mod * a % mod) % mod;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
const long long int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
const int maxm = 2e3 + 50;
const double eps = 1e-8;
const long long int inf = 0x3f3f3f3f;
const long long int lnf = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1);
int n, m;
long long int a[maxn];
string s;
void sove() {
  long long int mi = inf;
  long long int ma = 0;
  cin >> n;
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (i > 1) ans = max(ans, a[i] * a[i - 1]);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    sove();
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long int ksm(long long int a, long long int b, long long int mod) {
  long long int ans = 1;
  while (b) {
    if (b & 1) {
      ans = (a % mod * ans % mod) % mod;
    }
    b >>= 1;
    a = (a % mod * a % mod) % mod;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
const long long int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
const int maxm = 2e3 + 50;
const double eps = 1e-8;
const long long int inf = 0x3f3f3f3f;
const long long int lnf = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1);
int n, m;
long long int a[maxn];
string s;
void sove() {
  long long int mi = inf;
  long long int ma = 0;
  cin >> n;
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (i > 1) ans = max(ans, a[i] * a[i - 1]);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    sove();
  }
  return 0;
}

```