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;
  using namespace std::chrono;
  using ll = long long;
  using db = long double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using pd = pair<db, db>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vd = vector<db>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  using vpd = vector<pd>;
  const int MX = 2e5 + 5;
  const int MOD = 1e9 + 7;
  const ll INF = 1e18;
  const db PI = acos((db)-1);
  const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
  mt19937 rng((uint32_t)steady_clock::now().time_since_epoch().count());
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T, class func>
  T first_true(T lo, T hi, func first) {
    hi++;
    assert(lo <= hi);
    while (lo < hi) {
      T mid = lo + (hi - lo) / 2;
      first(mid) ? hi = mid : lo = mid + 1;
    }
    return lo;
  }
  template <class T, class func>
  T last_true(T lo, T hi, func first) {
    lo--;
    assert(lo <= hi);
    while (lo < hi) {
      T mid = lo + (hi - lo + 1) / 2;
      first(mid) ? lo = mid : hi = mid - 1;
    }
    return lo;
  }
  template <class T>
  void remove_duplicates(vector<T>& vec) {
    sort((vec).begin(), (vec).end());
    vec.erase(unique((vec).begin(), (vec).end()), end(vec));
  }
  void setIO(str S = "") {
    cin.tie(nullptr)->sync_with_stdio(false);
    cout << fixed << setprecision(15);
    if ((int)((S).size()) > 0) {
      freopen(S.c_str(), "r", stdin);
      freopen(S.c_str(), "w", stdout);
    }
  }
  void run_case(int tc = 0) {
    int n, k;
    cin >> n >> k;
    vl arr(n);
    for (auto& x : arr) cin >> x;
    ll res = -INF;
    for (int i = n - 1; i >= 0; --i) {
      if (res > (ll)i * (i + 1)) break;
      for (int j = i - 1; (ll)(i + 1) * (j + 1) > res && j >= 0; --j) {
        res = max(res, (ll)(i + 1) * (j + 1) - k * (arr[i] | arr[j]));
      }
    }
    cout << res << "\n";
  }
  int32_t main() {
    setIO();
    int tc = 1;
    cin >> tc;
    for (int t = 0; t < tc; ++t) run_case(t);
  }
  