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 _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T>
  void _print(stack<T> v);
  template <class T>
  void _print(list<T> v);
  template <class T>
  void _print(priority_queue<T> v);
  template <class T>
  void _print(priority_queue<T> v) {
    cerr << "{ ";
    while (!v.empty()) {
      _print(v.top());
      cerr << " ";
      v.pop();
    }
    cerr << " }";
  }
  template <class T>
  void _print(stack<T> v) {
    cerr << "[";
    while (!v.empty()) {
      _print(v.top());
      cerr << " ";
      v.pop();
    }
    cerr << "]";
  }
  template <class T>
  void _print(list<T> v) {
    cerr << "[";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  long long power(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) {
        res *= a;
      }
      b /= 2;
      a *= a;
    }
    return res;
  }
  long long mod_pow(long long a, long long b, long long mod) {
    long long res = 1;
    while (b) {
      if (b & 1) {
        res = (res * a) % mod;
      }
      b /= 2;
      a = (a * a) % mod;
    }
    return (res % mod);
  }
  void usaco(string filename = "") {
    if (((int)(filename).size())) {
      freopen((filename + ".in").c_str(), "r", stdin);
      freopen((filename + ".out").c_str(), "w", stdout);
    }
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n + 1, 0);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    long long ans = -1e18;
    for (int j = n; j >= max(n - 500, 1ll); --j) {
      for (int i = j - 1; i >= max(n - 500, 1ll); --i) {
        long long val = i * 1ll * j;
        val -= k * (a[i] | a[j]);
        ans = max(ans, val);
      }
    }
    cout << ans << "\n";
  }
  int main() {
    cout << setprecision(15) << fixed;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int test = 1, testcase = 1;
    cin >> test;
    while (test--) {
      solve();
    }
  }
  