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>
using namespace std;
typedef long long ll;

const int N = 55;
int a[N], n;
ll c[N][N], sc[N][N];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k, s, t;
    cin >> n >> k >> s >> t;
    for (int i = 0; i < 18; i++) if ((s & (1<<i)) == 1 && (t & (1<<i)) == 0) {
        cout << "0\n";
        return 0;
    }
    c[0][0] = 1;
    for (int i = 1; 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];
    }
    for (int i = 0; i <= n; ++i) {
        sc[i][0] = 0;
        for (int j = 1; j <= k; ++j) sc[i][j] = sc[i][j - 1] + c[i][j];
    }
    for (int i = 1; i <= n; ++i) cin >> a[i];
    {
        static int b[N];
        for (int i = 1; i <= n; ++i) b[i] = a[i];
        int tot = 0;
        for (int i = 1; i <= n; ++i) {
            bool flg = 1;
            for (int j = 0; j <18; ++j) {
                if ((s & (1<<j)) == (t & (1<<j)) && (a[i] & (1<<j)) != (s & (1<<j))) {
                    flg = false;
                    break;
                }
            }
            if (flg) b[++tot] = a[i];
        }
        n = tot;
        for (int i = 1; i <= n; ++i) a[i] = b[i];
    }
    ll ans = 0;
    for (int mask = s ^ t; ; mask = (mask - 1) & (s^t)) {
        static int cnt[1 << 19];
        for (int i = 1; i <= n; ++i) {
            ++cnt[a[i] & mask];
        }
        ll tmp = 0;
        for (int i = 1; i <= n; ++i) {
            tmp += sc[cnt[a[i]&mask]][k];
            cnt[a[i]&mask] = 0; 
        }
        if (__builtin_popcount(mask) & 1)
            ans -= tmp;
        else ans += tmp;
        if (!mask) break;
    }
    cout << ans << '\n';
    return 0;
}