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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
#define N 233
const int dig=18;
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,m,a[N],k,S,T;
ll ans,C[N][N];
ll calc(int x){
	static int cnt[1<<18],vis[1<<18];
	for(int i=1;i<=n;++i){
		++cnt[a[i]&x];
	}
	ll tot=0;
	for(int i=1;i<=n;++i){
		int c=a[i]&x;
		if(vis[c])continue;
		vis[c]=1;
		for(int j=1;j<=k;++j){
			tot+=C[cnt[c]][j];
		}
	}
	for(int i=1;i<=n;++i){
		vis[a[i]&x]=cnt[a[i]&x]=0;
	}
	return tot;
}
int main(){
	n=read(),k=read(),S=read(),T=read();
	for(int i=0;i<dig;++i){
		if((S>>i)&1&&!((T>>i)&1)){
			return !printf("0\n");
		}
	}
	for(int i=1;i<=n;++i){
		int x=read();
		bool ok=true;
		for(int j=0;j<dig;++j){
			if((S>>j)&1&&!((x>>j)&1)||!((T>>j)&1)&&(x>>j)&1){
				ok=false;break;
			}
		}
		if(ok)a[++m]=x;
	}
	C[0][0]=1;
	for(int i=1;i<=n;++i){
		C[i][0]=1;
		for(int j=1;j<=i;++j){
			C[i][j]=C[i-1][j-1]+C[i-1][j];
		}
	}
	if(!n)return !printf("0\n");
	n=m;
	for(int i=S^T;;i=(i-1)&(S^T)){
		ans+=__builtin_popcount(i)&1?-calc(i):calc(i);
		if(!i)break;
	}
	printf("%lld\n",ans);
	return 0;
}
