#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long MOD = 998244353;
const long long inf = 1e18;
const long long MAX = 2e5 + 1;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
inline long long ceil(long long a, long long b) { return (a + b - 1) / (b); }
vector<long long> sieve(int n) {
  int *arr = new int[n + 1]();
  vector<long long> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long pwr(long long x, long long n) {
  x = x % mod;
  if (!n) return 1;
  if (n & 1)
    return mul(x, pwr(mul(x, x), (n - 1) / 2));
  else
    return pwr(mul(x, x), n / 2);
}
long long pwrmod(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * pwrmod(b * b % m, e / 2, m) % m;
  return pwrmod(b * b % m, e / 2, m);
}
long long modinv(long long n) { return pwr(n, mod - 2); }
long long inv(long long i) {
  if (i == 1) return 1;
  return (MOD - (MOD / i) * inv(MOD % i) % MOD) % MOD;
}
struct cmp {
  bool operator()(const int &k1, const int &k2) { return k1 > k2; }
};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
template <typename T1, typename T2>
using safe_map = unordered_map<T1, T2, custom_hash>;
template <typename T1>
using safe_set = unordered_set<T1, custom_hash>;
void solve() {
  long long N;
  cin >> N;
  vector<long long> arr(N, 0);
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
  }
  long long highest = arr[0];
  long long highestIndex = 0;
  if (N == 1) {
    cout << highest << "\n";
    return;
  }
  long long mx = 0;
  for (int i = 1; i < N; i++) {
    mx = max(mx, arr[i] * arr[i - 1]);
  }
  cout << mx << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TEST = 1;
  cin >> TEST;
  for (long long tt = 1; tt <= TEST; tt++) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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;
const long long mod = 1000000007;
const long long MOD = 998244353;
const long long inf = 1e18;
const long long MAX = 2e5 + 1;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
inline long long ceil(long long a, long long b) { return (a + b - 1) / (b); }
vector<long long> sieve(int n) {
  int *arr = new int[n + 1]();
  vector<long long> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long pwr(long long x, long long n) {
  x = x % mod;
  if (!n) return 1;
  if (n & 1)
    return mul(x, pwr(mul(x, x), (n - 1) / 2));
  else
    return pwr(mul(x, x), n / 2);
}
long long pwrmod(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * pwrmod(b * b % m, e / 2, m) % m;
  return pwrmod(b * b % m, e / 2, m);
}
long long modinv(long long n) { return pwr(n, mod - 2); }
long long inv(long long i) {
  if (i == 1) return 1;
  return (MOD - (MOD / i) * inv(MOD % i) % MOD) % MOD;
}
struct cmp {
  bool operator()(const int &k1, const int &k2) { return k1 > k2; }
};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
template <typename T1, typename T2>
using safe_map = unordered_map<T1, T2, custom_hash>;
template <typename T1>
using safe_set = unordered_set<T1, custom_hash>;
void solve() {
  long long N;
  cin >> N;
  vector<long long> arr(N, 0);
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
  }
  long long highest = arr[0];
  long long highestIndex = 0;
  if (N == 1) {
    cout << highest << "\n";
    return;
  }
  long long mx = 0;
  for (int i = 1; i < N; i++) {
    mx = max(mx, arr[i] * arr[i - 1]);
  }
  cout << mx << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TEST = 1;
  cin >> TEST;
  for (long long tt = 1; tt <= TEST; tt++) {
    solve();
  }
  return 0;
}

```