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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e2 + 10;
int a[maxn], n, k, s, t, tot;
int vis[300010];
ll C[51][51], D[51][51];
int main() {
	scanf("%d%d%d%d",&n,&k,&s,&t);
	
	for (int i = 0; i <= n; i++)
		D[i][0] = C[i][0] = 1;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= i; j++) {
			C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
			D[i][j] = D[i][j - 1] + C[i][j]; 
		}
	for (int i = 1, x; i <= n; i++) {
		scanf("%d",&x);
		if ((x & s) == s && (x & t) == x)
			a[++tot] = x;
	} 
	t ^= s;
	int cnt = 0;
	for (int j = 0; j < 18; j++)
		if (t >> j & 1) cnt++;
	//printf("%d %d\n",cnt,tot);
	for (int i = 1; i <= tot; i++) {
		int val = 1, sum = 0;
		for (int j = 0; j < 18; j++) {
			if ((a[i] >> j & 1) && (t >> j & 1))
				sum += val;
			if (t >> j & 1)
				val *= 2;
		}
		a[i] = sum;
	}
	ll ans = 0;
	for (int i = 0; i < (1 << cnt); i++)
		vis[i] = 0;
	for (int t = 0; t < (1 << cnt); t++) {
		int s = 1;
		for (int j = 0; j < cnt; j++)
			if (t >> j & 1) s *= -1;
		ll sum = 0;
		for (int i = 1; i <= tot; i++) {
			sum -= D[vis[a[i] & t]][min(vis[a[i] & t],k)];
			vis[a[i] & t]++;
			sum += D[vis[a[i] & t]][min(vis[a[i] & t],k)];
		}
		//if (t == 0) printf("****%lld %lld\n",vis[0],C[5][5]);
		for (int i = 1; i <= tot; i++)
			vis[a[i] & t]--;
		//printf("%lld %lld\n",t,s * sum);
		ans += s * sum;
	}
	printf("%lld\n",ans);
	return 0;
}