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;
  template <class T, class U>
  inline bool chmax(T& a, U b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T, class U>
  inline bool chmin(T& a, U b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  constexpr int INF = 1000000000;
  constexpr ll llINF = 3000000000000000000;
  constexpr int mod = 998244353;
  constexpr double eps = 1e-10;
  const double pi = acos(-1);
  int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
  int Random(int mi, int ma) {
    random_device rnd;
    mt19937 mt(rnd());
    uniform_int_distribution<int> engine(mi, ma);
    return engine(mt);
  }
  bool prime(int a) {
    if (a == 1) return false;
    for (int i = 2; i * i <= a; i++) {
      if (a % i == 0) return false;
    }
    return true;
  }
  vector<int> primes;
  void init_prime(int n) {
    primes.push_back(2);
    for (int i = 3; i <= n; i += 2) {
      bool f = true;
      for (int j : primes) {
        if (j * j > i) break;
        if (i % j == 0) {
          f = false;
          break;
        }
      }
      if (f) primes.push_back(i);
    }
  }
  ll modpow(ll a, ll b) {
    ll res = 1;
    while (b) {
      if (b & 1) {
        res *= a;
        res %= mod;
      }
      a *= a;
      a %= mod;
      b >>= 1;
    }
    return res;
  }
  vector<ll> inv, fact, factinv;
  void init_fact(int n) {
    inv.resize(n + 1);
    fact.resize(n + 1);
    factinv.resize(n + 1);
    inv[0] = 0;
    inv[1] = 1;
    fact[0] = 1;
    factinv[0] = 1;
    for (ll i = 1; i <= n; i++) {
      if (i >= 2) inv[i] = mod - ((mod / i) * inv[mod % i] % mod);
      fact[i] = (fact[i - 1] * i) % mod;
      factinv[i] = factinv[i - 1] * inv[i] % mod;
    }
  }
  ll _inv(ll a, ll m = mod) {
    ll b = m, u = 1, v = 0;
    while (b) {
      ll t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
  }
  ll comb(int a, int b) {
    if (a < b) return 0;
    if (a < 0) return 0;
    return fact[a] * factinv[a - b] % mod * factinv[b] % mod;
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    ll ans = -llINF;
    for (int i = max(0, n - k - k); i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        chmax(ans, 1LL * (i + 1) * (j + 1) - k * (a[i] | a[j]));
      }
    }
    cout << ans << endl;
  }
  int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) solve();
  }
  