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;
  void swap(long long &x, long long &y) {
    long long temp = x;
    x = y;
    y = temp;
  }
  void google(long long t) { cout << "Case #" << t << ": "; }
  long long mod_add(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long mod_mul(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long mod_sub(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long expom(long long x, long long y, long long m) {
    if (y == 0) return 1;
    x = x % m;
    if (y % 2 == 0) return expom(x * x, y / 2, m) % m;
    return (x * expom(x * x, y / 2, m)) % m;
  }
  inline long long gcd(long long a, long long b) {
    return b == 0 ? a : gcd(b, a % b);
  }
  inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
  inline long long ceil(long long a, long long b) {
    return (a % b == 0) ? a / b : a / b + 1;
  }
  long long dx[] = {1, 0, -1, 0};
  long long dy[] = {0, 1, 0, -1};
  vector<long long> pri(200000 + 6);
  void prime_factorization_sieve() {
    pri[1] = 1;
    for (long long i = 2; i <= 200000 + 5; i++) {
      pri[i] = -1;
    }
    for (long long i = 2; i <= 200000 + 5; i++) {
      if (pri[i] == -1) {
        for (long long j = i; j <= 200000 + 5; j += i) {
          if (pri[j] == -1) pri[j] = i;
        }
      }
    }
  }
  vector<long long> get_factors(long long n) {
    vector<long long> res;
    while (pri[n] != 1) {
      res.push_back(pri[n]);
      n = n / pri[n];
    }
    return res;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> arr(n + 1);
    for (long long i = 1; i <= n; i++) cin >> arr[i];
    long long ans = INT_MIN;
    long long x = (n >= 200) ? n - 200 : n;
    if (n >= 200) {
      for (long long i = x + 1; i <= n; i++) {
        for (long long j = 1; j <= n; j++) {
          if (i != j) {
            ans = ((ans) < (i * j - k * (arr[i] | arr[j]))
                       ? (i * j - k * (arr[i] | arr[j]))
                       : (ans));
          }
        }
      }
    } else {
      for (long long i = 1; i <= x; i++) {
        for (long long j = 1; j <= n; j++) {
          if (i != j) {
            ans = ((ans) < (i * j - k * (arr[i] | arr[j]))
                       ? (i * j - k * (arr[i] | arr[j]))
                       : (ans));
          }
        }
      }
    }
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  