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;
const int MN = 100005;
const int MX = 500005;
const int INF = 1000000007;
const int MOD = 998244353;
int n, k;
int ma[500005], a[500005];
set<int> s[500005];
bool f[500005];
inline int blen(int x) { return 32 - __builtin_clz(x); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int Tc;
  for (cin >> Tc; Tc--;) {
    cin >> n >> k;
    for (int i = 0; i < 4 * n + 1; i++) f[i] = 0, ma[i] = 0, s[i].clear();
    for (int i = 1; i < n + 1; i++)
      cin >> a[i], ma[a[i]] = i, s[a[i]].insert(i);
    int fir = 0, sec = 0;
    long long res = -1e9;
    for (int i = 0; i <= n; i++) {
      if (s[i].size() >= 2) {
        auto it = --s[i].end();
        fir = *it;
        it--;
        sec = *it;
        res = max(res, 1LL * fir * sec - k * i);
      }
    }
    int l = blen(n);
    int ii, jj;
    for (int i = (1 << l) - 1; i >= 1; i--) {
      ii = 0;
      jj = 0;
      for (int j = i; j; j = (j - 1) & i) {
        if (ii < ma[j]) {
          if (ii)
            jj = ii;
          else
            ii = ma[j];
        } else if (jj < ma[j])
          jj = ma[j];
      }
      if (ii && jj) res = max(res, 1LL * ii * jj - k * i);
    }
    cout << res << endl;
  }
  return 0;
}
