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>
#define ll long long
using namespace std;
const int MAXN = 51;
const int MAXV = 1 << 18;
int a[MAXN], b[MAXN];
int cnt[MAXV];
ll dp[MAXV], C[MAXN][MAXN];
int main() {
    for (int i = 0; i <= 50; 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 = 1; i <= 50; i++)
        for (int j = 1; j <= i; j++) C[i][j] += C[i][j - 1];

    int n, k, S, T; scanf("%d %d %d %d", &n, &k, &S, &T);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int bit = 0; bit < 18; bit++) {
        int s = (S >> bit & 1), t = (T >> bit & 1);
        if (s == 1 && t == 0) {
            puts("0");
            return 0;
        }
        for (int i = 1; i <= n; i++) {
            int x = (a[i] >> bit & 1);
            if (s == t && x != s) {
                swap(a[i], a[n]);
                n--;
                i--;
            }
        }
    }
    int bc = 0;
    for (int bit = 0; bit < 18; bit++) {
        int s = (S >> bit & 1), t = (T >> bit & 1);
        if (s == 0 && t == 1) {
            for (int i = 1; i <= n; i++) b[i] |= (a[i] >> bit & 1) * (1 << bc);
            bc++;
        }
    }
    ll ans = 0;
    vector<int>vec;
    for (int s = 0; s < (1 << bc); s++) {
        vec.clear();
        for (int i = 1; i <= n; i++) {
            if (!cnt[s & b[i]]) vec.push_back(s & b[i]);
            cnt[s & b[i]]++;
        }
        for (int x : vec) {
            dp[s] += C[cnt[x]][min(cnt[x], k)] - 1;
            cnt[x] = 0;
        }
        __builtin_popcount(s) & 1 ? ans -= dp[s] : ans += dp[s];
    }
    printf("%lld\n", ans);
    return 0;
}