#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;
}

### Prompt

In cpp, your task is to solve the following problem:

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.

### Response

```cpp
#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;
}

```