#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
void debug() { cerr << '\n'; }
template <typename T, typename... Ts>
void debug(T x, Ts... y) {
  cerr << "\033[31m" << x << "\033[0m";
  if (sizeof...(y) > 0) cerr << ' ';
  debug(y...);
}
template <typename T>
void debug(const T &a, int l, int r, char c) {
  for (int i = l; i <= r; ++i)
    cerr << "\033[31m" << a[i] << "\033[0m" << (i == r ? '\n' : c);
}
template <typename T>
void debug(vector<T> a) {
  for (int i = 0; i < (int)a.size(); ++i)
    cerr << "\033[31m" << a[i] << "\033[31m"
         << " \n"[i == ((int)a.size() - 1)];
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long qpow(long long x, long long y) {
  long long ans = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) ans *= x;
    x *= x;
  }
  return ans;
}
long long qpow(long long x, long long y, int MOD) {
  long long ans = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) ans = ans * x % MOD;
    x = x * x % MOD;
  }
  return ans;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  long long X = x;
  x = y;
  y = X - a / b * y;
}
long long inv(long long x, int mod = MOD) { return qpow(x, mod - 2, mod); }
long long m_m(long long a) { return (a % MOD + MOD) % MOD; }
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto &it : a) cin >> it;
    long long ma = 0;
    for (int i = (1); i < (n); ++i) ma = max(ma, 1ll * a[i] * a[i - 1]);
    cout << ma << '\n';
  }
  return 0;
}

### 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>
#pragma GCC optimize(2)
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
void debug() { cerr << '\n'; }
template <typename T, typename... Ts>
void debug(T x, Ts... y) {
  cerr << "\033[31m" << x << "\033[0m";
  if (sizeof...(y) > 0) cerr << ' ';
  debug(y...);
}
template <typename T>
void debug(const T &a, int l, int r, char c) {
  for (int i = l; i <= r; ++i)
    cerr << "\033[31m" << a[i] << "\033[0m" << (i == r ? '\n' : c);
}
template <typename T>
void debug(vector<T> a) {
  for (int i = 0; i < (int)a.size(); ++i)
    cerr << "\033[31m" << a[i] << "\033[31m"
         << " \n"[i == ((int)a.size() - 1)];
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long qpow(long long x, long long y) {
  long long ans = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) ans *= x;
    x *= x;
  }
  return ans;
}
long long qpow(long long x, long long y, int MOD) {
  long long ans = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) ans = ans * x % MOD;
    x = x * x % MOD;
  }
  return ans;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  long long X = x;
  x = y;
  y = X - a / b * y;
}
long long inv(long long x, int mod = MOD) { return qpow(x, mod - 2, mod); }
long long m_m(long long a) { return (a % MOD + MOD) % MOD; }
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto &it : a) cin >> it;
    long long ma = 0;
    for (int i = (1); i < (n); ++i) ma = max(ma, 1ll * a[i] * a[i - 1]);
    cout << ma << '\n';
  }
  return 0;
}

```