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 <cstdio>

typedef long long LL;
const int MN = 55, MV = 1 << 18;

LL Binom[MN][MN], Ans;
int N, K, S, T, A[MN], part[MV];

int main() {
	scanf("%d%d%d%d", &N, &K, &S, &T);
	for (int i = 1; i <= N; ++i) {
		scanf("%d", &A[i]);
		if ((A[i] & S) != S || (A[i] | T) != T) {
			--i, --N;
			continue;
		}
	}
	for (int i = 0; i <= N; ++i) {
		Binom[i][0] = 1;
		for (int j = 1; j <= i; ++j)
			Binom[i][j] = Binom[i - 1][j - 1] + Binom[i - 1][j];
	}
	for (int i = 0; i <= N; ++i)
		for (int j = 2; j <= K; ++j)
			Binom[i][j] += Binom[i][j - 1];
	for (int X = 0; X < MV; ++X) {
		if ((X & S) != S || (X | T) != T) continue;
		LL Sum = 0;
		for (int i = 1; i <= N; ++i) ++part[A[i] & X];
		for (int i = 1; i <= N; ++i) if(part[A[i] & X])
			Sum += Binom[part[A[i] & X]][K], part[A[i] & X] = 0;
		Ans += Sum * (__builtin_parity(X ^ S) ? -1 : 1);
	}
	printf("%lld\n", Ans);
	return 0;
}