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;
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t;
    cin>>t;
    while(t--){
        long long int n,k;
        cin>>n>>k;
        long long int arr[n+1]={0};
        for(int i=1;i<=n;i++){
            cin>>arr[i];
        }
        long long int aor[n]={0};
        for(int i=1;i<n;i++){
            aor[i]=arr[i]|arr[i+1];
        }
        long long int ans1=(n*(n-1))-(k*aor[n-1]);
        long long int ans=ans1;
        long long int temp;
        for(long long int i=n-2;i>0;i--){
            if((i*(i+1))<ans1){
                break;
            }
            temp=(i*(i+1))-(k*aor[i]);
            if(temp>ans){
                ans=temp;
            }
        }
        cout<<ans<<endl;
    }    
    return 0;
}  
                          