#include <bits/stdc++.h>
using namespace std;
void setIO(string name = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (int((name).size())) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long mcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool prime(long long n, long long i) {
  if (i > sqrt(n)) return true;
  if (n % i == 0)
    return false;
  else
    return prime(n, i + 1);
}
struct compii {
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    if (a.first == a.second) return a.second < b.second;
    return a.first > b.first;
  }
};
bool comp(int a, int b) { return a > b; }
namespace operators {
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
  in >> x.first >> x.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> x) {
  out << x.first << " " << x.second;
  return out;
}
template <typename T1>
istream& operator>>(istream& in, vector<T1>& x) {
  for (auto& i : x) in >> i;
  return in;
}
template <typename T1>
ostream& operator<<(ostream& out, vector<T1>& x) {
  for (auto& i : x) out << i << " ";
  return out;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, vector<pair<T1, T2>>& x) {
  for (auto& i : x) out << i.first << " " << i.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, vector<pair<T1, T2>>& x) {
  for (auto& i : x) in >> i.first >> i.second;
  return in;
}
}  // namespace operators
using namespace operators;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int mod = 1e9 + 7;
const long long inf = 1e18;
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  if (n == 2) {
    cout << a[1 - 1] * a[2 - 1] << endl;
    return;
  }
  long long ans = 0;
  for (int i = 1; i < n - 1; i++) {
    ans = max(ans, a[i] * a[i - 1]);
    ans = max(ans, a[i] * a[i + 1]);
  }
  cout << ans << endl;
}
int main() {
  setIO("");
  int t = 1;
  cin >> t;
  while (t-- > 0) solve();
  return 0;
}

### Prompt

Your task is to create 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;
void setIO(string name = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (int((name).size())) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long mcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool prime(long long n, long long i) {
  if (i > sqrt(n)) return true;
  if (n % i == 0)
    return false;
  else
    return prime(n, i + 1);
}
struct compii {
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    if (a.first == a.second) return a.second < b.second;
    return a.first > b.first;
  }
};
bool comp(int a, int b) { return a > b; }
namespace operators {
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
  in >> x.first >> x.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> x) {
  out << x.first << " " << x.second;
  return out;
}
template <typename T1>
istream& operator>>(istream& in, vector<T1>& x) {
  for (auto& i : x) in >> i;
  return in;
}
template <typename T1>
ostream& operator<<(ostream& out, vector<T1>& x) {
  for (auto& i : x) out << i << " ";
  return out;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, vector<pair<T1, T2>>& x) {
  for (auto& i : x) out << i.first << " " << i.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, vector<pair<T1, T2>>& x) {
  for (auto& i : x) in >> i.first >> i.second;
  return in;
}
}  // namespace operators
using namespace operators;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int mod = 1e9 + 7;
const long long inf = 1e18;
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  if (n == 2) {
    cout << a[1 - 1] * a[2 - 1] << endl;
    return;
  }
  long long ans = 0;
  for (int i = 1; i < n - 1; i++) {
    ans = max(ans, a[i] * a[i - 1]);
    ans = max(ans, a[i] * a[i + 1]);
  }
  cout << ans << endl;
}
int main() {
  setIO("");
  int t = 1;
  cin >> t;
  while (t-- > 0) solve();
  return 0;
}

```