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>
#include <vector>
using namespace std;
int a[60];
vector<int> v;
long long dp[2][51][1<<18];
int main() {
	int n, k, s, t;
	scanf("%d%d%d%d", &n, &k, &s, &t);
	if ((s & t) != s) {
		puts("0");
		return 0;
	}
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = 0; i < n; i++) {
		if ((a[i] & s) == s && (a[i] | t) == t) {
			v.push_back(a[i]);
		}
	}
	int ma = (1 << 18) - 1;
	long long ans = 0;
	for (int i = 0; i < v.size(); i++) {
		dp[0][i][0] = 1;
	}
	int y = 0, z = 1;
	for (int i = 0; i < k; i++) {
		for (int j = 0; j < v.size(); j++) {
			for (int l = 0; l < 1 << 18; l++) {
				dp[z][j][l] = 0;
			}
		}
		for (int j = 0; j < v.size(); j++) {
			for (int l = 0; l < 1 << 18; l++) {
				if (dp[y][j][l] == 0) continue;
				for (int m = j + 1; m < v.size(); m++) {
					int o = v[j] & ~l, e = (ma ^ v[j]) & ~l;
					int no = o & v[m], ne = e & ~v[m];
					int nl = ma ^ no ^ ne;
					dp[z][m][nl] += dp[y][j][l];
				}
			}
			ans += dp[y][j][s^t];
		}
		for (int j = 0; j < v.size(); j++) {
			for (int l = 0; l < 1 << 18; l++) {
				dp[y][j][l] = dp[z][j][l];
			}
		}
		y = 1 - y;
		z = 1 - z;
	}
	printf("%lld\n", ans);
	return 0;
}