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 ll = long long;
  using ld = long double;
  template <class T>
  using mpq = priority_queue<T, vector<T>, greater<T>>;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int pct(const ll &x) { return __builtin_popcountll(x); }
  int bits(const ll &x) { return 63 - __builtin_clzll(x); }
  ll cdiv(const ll &x, const ll &y) { return (x / y) + !!(x % y); }
  template <class T>
  bool chmin(T &a, const T &b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool chmax(T &a, const T &b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T>
  T uid(const T &a, const T &b) {
    return uniform_int_distribution<T>(a, b)(rng);
  }
  template <class T>
  bool inc(const T &a, const T &b, const T &c) {
    return a <= b && b <= c;
  }
  template <class T>
  int lb(const vector<T> &v, const T &a) {
    return lower_bound((v).begin(), (v).end(), a) - v.begin();
  }
  template <class T>
  int ub(const vector<T> &v, const T &a) {
    return upper_bound((v).begin(), (v).end(), a) - v.begin();
  }
  template <class T>
  int mx(const vector<T> &v) {
    return max_element((v).begin(), (v).end()) - v.begin();
  }
  template <class T>
  int mn(const vector<T> &v) {
    return min_element((v).begin(), (v).end()) - v.begin();
  }
  template <class T>
  T sum(const vector<T> &v) {
    return accumulate((v).begin(), (v).end(), (T)0);
  }
  void USACO(string s = "") {
    if (((int)(s).size())) ifstream(s + ".in"), ofstream(s + ".out");
  }
  const int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
  const char dir[4] = {'R', 'L', 'D', 'U'}, ed = '\n';
  const ll INF = 1e18, MOD = 1e9 + 7, mxN = 1e5 + 5;
  void solve(int tc) {
    ll n, k, ans = -INF;
    cin >> n >> k;
    vector<ll> a(n);
    for (auto &i : a) cin >> i;
    for (ll i = 1; i <= n; ++i)
      for (ll j = i - 1; j >= max((ll)1, (ll)i - 200); --j)
        chmax(ans, i * j - k * (a[i - 1] | a[j - 1]));
    cout << ans << ed;
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(15);
    USACO();
    int t = 1;
    cin >> t;
    for (int tc = 0; tc < t; ++tc) solve(tc);
    return 0;
  }
  