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;
vector<long long> v[100005];
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    long long n, k;
    cin >> n >> k;
    long long ar[n];
    int d = 0;
    for (int i = 0; i < n; i++) {
      cin >> ar[i];
      v[ar[i]].push_back(i + 1);
    }
    long long mx = -1e9;
    for (int i = 0; i < n; i++) {
      if (v[ar[i]].size() >= 2) {
        long long p = v[ar[i]].back();
        v[ar[i]].pop_back();
        long long q = v[ar[i]].back();
        long long ans = (p * q) - (k * ar[i]);
        mx = max(mx, ans);
      }
    }
    v[ar[0]].clear();
    for (int i = 1; i < n; i++) {
      v[ar[i]].clear();
      long long ans = (i * (i + 1)) - (k * (ar[i] | ar[i - 1]));
      mx = max(mx, ans);
    }
    cout << mx << endl;
  }
}
