Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 55;
int n, K, S, T, a[maxn], b[maxn];
ll ans, C[maxn][maxn], coef[maxn], cnt[1 << 18];

int main() {
    scanf("%d %d %d %d", &n, &K, &S, &T);
    auto chk = [&](int x, int y) { return (x & y) == y; };
    if (!chk(T, S)) printf("0\n"), exit(0);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        if (!chk(a[i], S) || !chk(T, a[i])) { i--, n--; continue; }
    }
    T ^= S;
    for (int i = 1; i <= n; i++) {
        a[i] ^= S;
        for (int j = 0; j < 18; j++) if (T >> j & 1) {
            if (!(a[i] >> j & 1)) b[i] |= 1 << j;
        }
    }
    auto FWT = [&](ll *a, int lim) {
        for (int i = 1; i < lim; i <<= 1) {
            for (int j = 0; j < lim; j += i << 1) {
                for (int k = 0; k < i; k++) {
                    a[j + k] += a[i + j + k];
                }
            }
        }
    };
    for (int i = 0; i <= n; i++) {
        for (int j = C[i][0] = 1; j <= i; j++) {
            C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
        for (int j = 0; j < K; j++) {
            coef[i] += C[i][j];
        }
    }
    for (int i = 1; i <= n; i++) {
        memset(cnt, 0, sizeof(cnt));
        for (int j = i + 1; j <= n; j++) {
            cnt[b[i] ^ a[j]]++;
        }
        FWT(cnt, 1 << 18);
        for (int j = 0; j < 1 << 18; j++) {
            if (__builtin_popcount(j) & 1) ans -= coef[cnt[j]];
            else ans += coef[cnt[j]];
        }
    }
    printf("%lld\n", ans);
    return 0;
}