#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e8;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
const int N = (int)1e5 + 10;
void TimeExecution() {
  cout << "\n\n===>> Time Execution: " << clock() / (double)1000 << " sec(s).";
}
long long sqr(long long a) { return ((a % MOD) * (a % MOD)) % MOD; }
long long msub(long long a, long long b) { return ((a - b) % MOD + MOD) % MOD; }
long long madd(long long a, long long b) { return (a + b) % MOD; }
long long mmul(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    pair<long long, long long> a[n + 1];
    long long ans = -LINF;
    for (int i = (1); i <= (n); i++) {
      long long val;
      cin >> val;
      a[i].first = val;
      a[i].second = i;
    }
    for (int i = (1); i <= (n - 1); i++)
      ans = max(ans, i * (i + 1) - k * (a[i].first | a[i + 1].first));
    for (int j = (1); j <= (k); j++)
      for (int i = (1); i <= (n - j); i++)
        ans = max(ans, a[i].second * a[i + j].second -
                           k * (a[i].first | a[i + j].first));
    sort(a + 1, a + 1 + n);
    for (int j = (1); j <= (k); j++)
      for (int i = (1); i <= (n - j); i++)
        ans = max(ans, a[i].second * a[i + j].second -
                           k * (a[i].first | a[i + j].first));
    cout << ans;
    cout << '\n';
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e8;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
const int N = (int)1e5 + 10;
void TimeExecution() {
  cout << "\n\n===>> Time Execution: " << clock() / (double)1000 << " sec(s).";
}
long long sqr(long long a) { return ((a % MOD) * (a % MOD)) % MOD; }
long long msub(long long a, long long b) { return ((a - b) % MOD + MOD) % MOD; }
long long madd(long long a, long long b) { return (a + b) % MOD; }
long long mmul(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    pair<long long, long long> a[n + 1];
    long long ans = -LINF;
    for (int i = (1); i <= (n); i++) {
      long long val;
      cin >> val;
      a[i].first = val;
      a[i].second = i;
    }
    for (int i = (1); i <= (n - 1); i++)
      ans = max(ans, i * (i + 1) - k * (a[i].first | a[i + 1].first));
    for (int j = (1); j <= (k); j++)
      for (int i = (1); i <= (n - j); i++)
        ans = max(ans, a[i].second * a[i + j].second -
                           k * (a[i].first | a[i + j].first));
    sort(a + 1, a + 1 + n);
    for (int j = (1); j <= (k); j++)
      for (int i = (1); i <= (n - j); i++)
        ans = max(ans, a[i].second * a[i + j].second -
                           k * (a[i].first | a[i + j].first));
    cout << ans;
    cout << '\n';
  }
  return 0;
}

```