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
#include <bits/stdc++.h>
#define long long long int
using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n, k, s, t;
    cin >> n >> k >> s >> t;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    vector<vector<long>> cc(n + 1, vector<long>(n + 1));
    for (int i = 0; i <= n; i++) {
        cc[i][0] = cc[i][i] = 1;
        for (int j = 1; j < i; j++) {
            cc[i][j] = cc[i - 1][j] + cc[i - 1][j - 1];
        }
    }
    for (int i = 0; i <= n; i++) {
//        cout << cc[i][0];
        for (int j = 1; j <= n; j++) {
            cc[i][j] += cc[i][j - 1];
//            cout << " " << cc[i][j];
        }
//        cout << "\n";
    }

    vector<int> aa;
    for (int x : a) {
        if ((x & s) == s && (x | t) == t) {
            aa.push_back(x);
        }
    }
    a = aa;
    n = a.size();

    long res = 0;
    for (int x = 0; x < (1 << 18); x++) {
        if ((s & x) != 0 || (t & x) != x) {
            continue;
        }
        long s = 0;
        for (int i = 0; i < n; i++) {
            int q = a[i] & x;
            int c = 0;
            for (int j = i + 1; j < n; j++) {
                if ((a[j] & x) == q) {
                    c++;
                }
            }
            s += cc[c][k - 1];
        }
//        cout << x << " " << s << "\n";
        if (__builtin_popcount(x) % 2 == 0) {
            res += s;
        } else {
            res -= s;
        }
    }
    cout << res << "\n";

    return 0;
}