Given are N pairwise distinct non-negative integers A_1,A_2,\ldots,A_N. Find the number of ways to choose a set of between 1 and K numbers (inclusive) from the given numbers so that the following two conditions are satisfied:

* The bitwise AND of the chosen numbers is S.
* The bitwise OR of the chosen numbers is T.

Constraints

* 1 \leq N \leq 50
* 1 \leq K \leq N
* 0 \leq A_i < 2^{18}
* 0 \leq S < 2^{18}
* 0 \leq T < 2^{18}
* A_i \neq A_j (1 \leq i < j \leq N)

Input

Input is given from Standard Input in the following format:


N K S T
A_1 A_2 ... A_N


Output

Print the answer.

Examples

Input

3 3 0 3
1 2 3


Output

2


Input

5 3 1 7
3 4 9 1 5


Output

2


Input

5 4 0 15
3 4 9 1 5


Output

3
/**
 *    author:  tourist
 *    created: 13.06.2020 15:18:47       
**/
#undef _GLIBCXX_DEBUG

#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k, s, t;
  cin >> n >> k >> s >> t;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if ((t & s) != s) {
    cout << 0 << '\n';
    return 0;
  }
  vector<vector<long long>> C(n + 1, vector<long long>(n + 1));
  for (int i = 0; i <= n; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
    }
  }
  vector<long long> ways(n + 1);
  for (int i = 0; i <= n; i++) {
    for (int j = 1; j <= min(i, k); j++) {
      ways[i] += C[i][j];
    }
  }
  vector<int> b;
  long long ans = 0;
  for (int mask = 0; mask < (1 << 18); mask++) {
    if ((mask & (t ^ s)) != mask) {
      continue;
    }
    int sign = (__builtin_popcount(mask) % 2 == 0 ? 1 : -1);
    b.clear();
    for (int i = 0; i < n; i++) {
      if ((a[i] & s) != s) {
        continue;
      }
      if ((a[i] | t) != t) {
        continue;
      }
      b.push_back(a[i] & mask);
    }
    sort(b.begin(), b.end());
    int beg = 0;
    while (beg < (int) b.size()) {
      int end = beg;
      while (end + 1 < (int) b.size() && b[end + 1] == b[end]) {
        ++end;
      }
      ans += sign * ways[end - beg + 1];
      beg = end + 1;
    }
  }
  cout << ans << '\n';
  return 0;
}
