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.
"""RANK1ZEN; 3966 PEAK; NA; FLEX SUPPORT: Zen, Bap; Battlenet ID -> Knuckles#11791"""
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# endregion ------------------------------------------------------------------------|
# region ---------------------------------------------------------------------------|
from sys import stdin, stdout
from bisect import bisect_left, bisect_right
from math import ceil, floor, log
from collections import Counter, deque
from heapq import heappush, heappop, heapify
from itertools import permutations, product
def re(): return stdin.readline().rstrip()
def mapi(): return map(int, re().split())
def veci(): return list(map(int, re().split()))
mod = 1000000007
# endregion ------------------------------------------------------------------------|

def solve(n, k, a):
    ans = -1e12
    for i in range(n, max(0, n - 200), -1):
        for j in range(i - 1, max(0, n - 200), -1):
            ans = max(ans, i * j - k * (a[i]|a[j]))
    return ans

for t in range(int(re())):
    n, k = mapi()
    a = [0] + veci()
    print(solve(n, k, a))
