#include <bits/stdc++.h>
using namespace std;
namespace LYYY {
inline int rd() {
  int s = 0;
  char c = getchar();
  int f = 1;
  while (c < 48 || c > 57) {
    if (c == 45) f = -1;
    c = getchar();
  }
  while (c >= 48 && c <= 57) s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
  return s * f;
}
long long rdl() {
  long long s = 0;
  char c = getchar();
  int f = 1;
  while (c < 48 || c > 57) {
    if (c == 45) f = -1;
    c = getchar();
  }
  while (c >= 48 && c <= 57) s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
  return s * f;
}
void wt(int x) {
  if (x >= 10) wt(x / 10);
  putchar(x % 10 + 48);
}
void WT(int x) {
  if (x < 0) x = -x, putchar(45);
  wt(x);
}
void wtl(long long x) {
  if (x >= 10) wtl(x / 10);
  putchar(x % 10 + 48);
}
void WTL(long long x) {
  if (x < 0) x = -x, putchar(45);
  wtl(x);
}
}  // namespace LYYY
using namespace LYYY;
int main() {
  int T = rd();
  while (T--) {
    int N = rd();
    long long ans = 0, tmp = rd();
    for (int i = 2; i <= N; ++i) {
      long long a = rd();
      ans = max(ans, a * tmp);
      tmp = a;
    }
    wtl(ans), puts("");
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
namespace LYYY {
inline int rd() {
  int s = 0;
  char c = getchar();
  int f = 1;
  while (c < 48 || c > 57) {
    if (c == 45) f = -1;
    c = getchar();
  }
  while (c >= 48 && c <= 57) s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
  return s * f;
}
long long rdl() {
  long long s = 0;
  char c = getchar();
  int f = 1;
  while (c < 48 || c > 57) {
    if (c == 45) f = -1;
    c = getchar();
  }
  while (c >= 48 && c <= 57) s = (s << 3) + (s << 1) + (c ^ 48), c = getchar();
  return s * f;
}
void wt(int x) {
  if (x >= 10) wt(x / 10);
  putchar(x % 10 + 48);
}
void WT(int x) {
  if (x < 0) x = -x, putchar(45);
  wt(x);
}
void wtl(long long x) {
  if (x >= 10) wtl(x / 10);
  putchar(x % 10 + 48);
}
void WTL(long long x) {
  if (x < 0) x = -x, putchar(45);
  wtl(x);
}
}  // namespace LYYY
using namespace LYYY;
int main() {
  int T = rd();
  while (T--) {
    int N = rd();
    long long ans = 0, tmp = rd();
    for (int i = 2; i <= N; ++i) {
      long long a = rd();
      ans = max(ans, a * tmp);
      tmp = a;
    }
    wtl(ans), puts("");
  }
  return 0;
}

```