You are given n integers a_1, a_2, …, a_n and an integer k. Find the maximum value of i ⋅ j - k ⋅ (a_i | a_j) over all pairs (i, j) of integers with 1 ≤ i < j ≤ n. Here, | is the [bitwise OR operator](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

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 two integers n (2 ≤ n ≤ 10^5) and k (1 ≤ k ≤ min(n, 100)).

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n).

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 i ⋅ j - k ⋅ (a_i | a_j).

Example

Input


4
3 3
1 1 3
2 2
1 2
4 3
0 1 2 3
6 6
3 2 0 0 5 6


Output


-1
-4
3
12

Note

Let f(i, j) = i ⋅ j - k ⋅ (a_i | a_j).

In the first test case, 

  * f(1, 2) = 1 ⋅ 2 - k ⋅ (a_1 | a_2) = 2 - 3 ⋅ (1 | 1) = -1. 
  * f(1, 3) = 1 ⋅ 3 - k ⋅ (a_1 | a_3) = 3 - 3 ⋅ (1 | 3) = -6. 
  * f(2, 3) = 2 ⋅ 3 - k ⋅ (a_2 | a_3) = 6 - 3 ⋅ (1 | 3) = -3. 



So the maximum is f(1, 2) = -1.

In the fourth test case, the maximum is f(3, 4) = 12.
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const double PI = acos(-1);
long long modulo(long long _a, long long _temp_mod = mod) {
  if (abs(_a) >= _temp_mod) {
    _a %= _temp_mod;
  }
  if (_a < 0) _a += _temp_mod;
  return _a;
}
long long mult(long long _a, long long _b, long long _temp_mod = mod) {
  long long _res = modulo(_a, _temp_mod) * modulo(_b, _temp_mod);
  return modulo(_res, _temp_mod);
}
long long add(long long _a, long long _b, long long _temp_mod = mod) {
  long long _res = modulo(_a, _temp_mod) + modulo(_b, _temp_mod);
  return modulo(_res, _temp_mod);
}
long long power(long long _a, long long _b, long long _temp_mod = mod) {
  if (_b < 0) return 0;
  long long _res = 1;
  while (_b) {
    if (_b & 1) _res = mult(_res, _a, _temp_mod);
    _a = mult(_a, _a, _temp_mod);
    _b >>= 1;
  }
  return _res;
}
void pre_comp(void) {}
void solve(long long _test) {
  long long n, k;
  cin >> n >> k;
  vector<long long> arr(n);
  long long ans = INT_MIN;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  if (n > 100) {
    ans = n * (n - 1) - k * (arr[n - 1] | arr[n - 2]);
  } else {
    for (long long i = 1; i <= n; i++) {
      for (long long j = i + 1; j <= n; j++) {
        long long temp = i * j - k * (arr[i - 1] | arr[j - 1]);
        ans = max(ans, temp);
      }
    }
  }
  cout << ans << '\n';
}
int32_t main(void) {
  std::ios::sync_with_stdio(0);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  pre_comp();
  long long _t = 1;
  cin >> _t;
  for (long long i = 1; i <= _t; i++) {
    solve(i);
  }
  return 0;
}
