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.

//Bismillahir Rahmanir Raheem
#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#define sf(x) scanf("%d",&x)
#define sfl(x) scanf("%lld",&x)
#define lli long long int
#define ll64 int64_t
#define pb push_back
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define frr(i,a) for(int i=0;i<a;i++)
#define frl(i,a) for(lli i=0;i<a;i++)
#define fr1(i,a) for(int i=1;i<=a;i++)
#define iter(x)  x.begin(),x.end()
#define Memset(a,x,n) for(int i=0;i<n;i++)a[i]=x;
#define fi first
#define si second
//std::string s=std::bitset<M>(n).to_string();  //number to binary string
//lower_bound returns an iterator pointing to the first element in the range [first,last) which has a value not less than ‘val’.
//upper_bound returns an iterator pointing to the first element in the range [first,last) which has a value greater than ‘val’.
//if mod is prime modInv(a,mod)==binExp(a,mod-2);
typedef pair<lli,lli>pll;
typedef tree<pair<pll,lli>,null_type,less<pair<pll,lli>>,rb_tree_tag,tree_order_statistics_node_update>indexed_set;
//if we are working with pair write pii in place of int.
//indexed_set s , declaration. s.insert(x) ,to insert x. find_by_order(x) , to find the xth smallest element in the set(starting from 0).  order_of_key(x) to find the lower bound position of key x.(starting from 0)

const lli inf=1e11;
void solve()
{
    lli n,m;
    cin>>n>>m;
    lli a[n];

    frr(i,n)cin>>a[i];
    lli ans=-inf;
    for(int i=n-1,j=0;j<min(100ll,n);i--,j++)
    {
        for(int k=0;k<n;k++)
        {
            if(i!=k)
            {
                lli vv=((i+1)*(k+1)-m*(a[i]|a[k]));
                ans=max(ans, vv);
            }
        }
    }
    cout<<ans<<endl;

}
int main()
{
    int t=1,tc=0;
    cin>>t;
    while(t--)
    {
        tc++;
        solve();
    }
}

