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.
import sys
import math
import heapq
from collections import defaultdict as dd
from collections import OrderedDict as od
from collections import deque
from itertools import permutations as pp
from itertools import combinations as cc
from sys import stdin
from functools import cmp_to_key
from functools import lru_cache
from bisect import bisect_left as bs_l
from bisect import bisect_right as bs_r
input = sys.stdin.readline
mod=10**9+7
sys.setrecursionlimit(10**5)               
T=int(input())
for _ in range(T):
    n,k=map(int,input().split())
    arr=list(map(int,input().split()))
    if n>1000:
        ans=-10**9
        brr=arr[::-1]
        for i in range(1000):
            for j in range(i+1,1000):
                tmp=(n-i)*(n-j)-k*(brr[i]|brr[j])
                ans=max(ans,tmp)
        print(ans)
    else:
        ans=-10**9
        for i in range(n):
            for j in range(i+1,n):
                tmp=(i+1)*(j+1)-k*(arr[i]|arr[j])
                ans=max(ans,tmp)
        print(ans)
