#include <bits/stdc++.h>
using namespace std;
vector<int> primes;
bool vis[31624 + 1];
int factorize(int n) {
  int count = 0;
  for (auto x : primes) {
    if (n < x) break;
    while (n % x == 0) {
      n = n / x;
      count++;
    }
  }
  if (n > 1) count++;
  return count;
}
void seive(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (!vis[i]) {
      for (int j = i * i; j <= n; j += i) {
        vis[j] = 1;
      }
    }
  }
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) primes.push_back(i);
  }
}
long long int calculatepower(long long int A, long long int B) {
  if (B == 0)
    return 1;
  else if (B % 2 == 0)
    return calculatepower((A * A) % 1000000007, B >> 1) % 1000000007;
  else
    return (A * calculatepower((A * A) % 1000000007, B >> 1) % 1000000007);
}
long long int fastExp(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b = b >> 1;
  }
  return res;
}
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 = INT_MIN;
  for (int i = 0; i < n - 1; i++) {
    ans = max(ans, a[i] * a[i + 1]);
  }
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<int> primes;
bool vis[31624 + 1];
int factorize(int n) {
  int count = 0;
  for (auto x : primes) {
    if (n < x) break;
    while (n % x == 0) {
      n = n / x;
      count++;
    }
  }
  if (n > 1) count++;
  return count;
}
void seive(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (!vis[i]) {
      for (int j = i * i; j <= n; j += i) {
        vis[j] = 1;
      }
    }
  }
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) primes.push_back(i);
  }
}
long long int calculatepower(long long int A, long long int B) {
  if (B == 0)
    return 1;
  else if (B % 2 == 0)
    return calculatepower((A * A) % 1000000007, B >> 1) % 1000000007;
  else
    return (A * calculatepower((A * A) % 1000000007, B >> 1) % 1000000007);
}
long long int fastExp(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b = b >> 1;
  }
  return res;
}
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 = INT_MIN;
  for (int i = 0; i < n - 1; i++) {
    ans = max(ans, a[i] * a[i + 1]);
  }
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```