description: 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.
solution: #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;
    long long ans = INT_MIN;
    for (long long i = max(1LL, n - 2 * k); i < n; i++) {
      for (long long j = i + 1; j <= n; j++) {
        ans = max(ans, i * j - k * (a[i - 1] | a[j - 1]));
      }
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t = 1;
    cin >> t;
    pre();
    while (t--) {
      solve();
    }
  }
  