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;

const int N = 55;

int a[N], frq[1 << 18];
long long C[N][N], sum[N][N];

inline int Count(int x) {
	if (x == 0) return 0;
	else return 32 - __builtin_clz(x);
}

int main() {
	
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
	
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	for (int i = 0; i < N; i++) {
		C[i][0] = 1, sum[i][0] = 0;
		for (int j = 1; j <= i; j++) {
			C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
		}
		for (int j = 1; j < N; j++) sum[i][j] = sum[i][j - 1] + C[i][j];
	}
	
	int n, k, S, T; cin >> n >> k >> S >> T;
	int an = 0;
	for (int i = 1; i <= n; i++) {
		int u; cin >> u;
		if ((u & S) != S || (u | T) != T) continue;
		int cur = 0;
		for (int j = 0; j < 18; j++) if (((~S) >> j & 1) && (T >> j & 1)) cur = cur << 1 | (u >> j & 1);
		a[++an] = cur;
	}
	n = an;
	long long ans = 0;
	int xx = __builtin_popcount(T) - __builtin_popcount(S);
	for (int s = 0; s < (1 << xx); s++) {
		for (int i = 1; i <= n; i++) {
			frq[a[i] & s]++;
		}
		int sgn = (__builtin_popcount(s) % 2 ? -1 : 1);
		for (int i = 1; i <= n; i++) if (frq[a[i] & s]) {
			ans += sgn * sum[frq[a[i] & s]][k];
			frq[a[i] & s] = 0;
		}
	}
	
	cout << ans << "\n";
	
	return 0;
}