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;
unsigned long long power(unsigned long long a, unsigned long long p,
                         unsigned long long mod = 1e9 + 7) {
  unsigned long long res = 1, run = p % mod;
  while (p > 0) {
    if (p & 1) {
      res = (res * run) % mod;
    }
    p = (p >> 1);
    run = (run * run) % mod;
  }
  return res;
}
unsigned long long inverse(unsigned long long a,
                           unsigned long long mod = 1e9 + 7) {
  return power(a, mod - 2, mod);
}
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void pre() {}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (auto &x : a) cin >> x;
  auto mycom = [](const vector<long long> &a, const vector<long long> &b) {
    return a[2] > b[2];
  };
  set<vector<long long>, decltype(mycom)> s(mycom);
  for (long long i = 2; i <= n; i++) {
    s.insert({i, i - 1, i * (i - 1)});
  }
  long long ans = INT_MIN;
  while (!s.empty()) {
    auto mx = *s.begin();
    long long i = mx[0], j = mx[1];
    if (i * j <= ans) break;
    ans = max(ans, i * j - k * (a[i - 1] | a[j - 1]));
    s.erase({i, j, i * j});
    if (j != 1) s.insert({i, j - 1, i * (j - 1)});
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  unsigned long long t = 1;
  cin >> t;
  pre();
  while (t--) {
    solve();
  }
}
