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 long long mod = 1000000007;
  const double PI = acos(-1);
  long long modulo(long long _a, long long _temp_mod = mod) {
    if (abs(_a) >= _temp_mod) {
      _a %= _temp_mod;
    }
    if (_a < 0) _a += _temp_mod;
    return _a;
  }
  long long mult(long long _a, long long _b, long long _temp_mod = mod) {
    long long _res = modulo(_a, _temp_mod) * modulo(_b, _temp_mod);
    return modulo(_res, _temp_mod);
  }
  long long add(long long _a, long long _b, long long _temp_mod = mod) {
    long long _res = modulo(_a, _temp_mod) + modulo(_b, _temp_mod);
    return modulo(_res, _temp_mod);
  }
  long long power(long long _a, long long _b, long long _temp_mod = mod) {
    if (_b < 0) return 0;
    long long _res = 1;
    while (_b) {
      if (_b & 1) _res = mult(_res, _a, _temp_mod);
      _a = mult(_a, _a, _temp_mod);
      _b >>= 1;
    }
    return _res;
  }
  void pre_comp(void) {}
  void solve(long long _test) {
    long long n, k;
    cin >> n >> k;
    vector<long long> arr(n);
    long long ans = INT_MIN;
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
    }
    for (long long i = 1; i <= n; i++) {
      for (long long j = i + 1; j - i < 110 && j <= n; j++) {
        long long temp = i * j - k * (arr[i - 1] | arr[j - 1]);
        ans = max(ans, temp);
      }
    }
    cout << ans << '\n';
  }
  int32_t main(void) {
    std::ios::sync_with_stdio(0);
    std::cin.tie(NULL);
    std::cout.tie(NULL);
    pre_comp();
    long long _t = 1;
    cin >> _t;
    for (long long i = 1; i <= _t; i++) {
      solve(i);
    }
    return 0;
  }
  