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;
  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;
  }
  