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;
typedef long long ll;
int n,K,S,T,a[55],i,j,ss,tt,b[55],m;
ll C[55][55],all,se[29][2],ans;
void dfs(int d,ll s,int v){
	if(d==m){
		b[__builtin_popcountll(s)]+=v;
		return;
	}
	dfs(d+1,s,v);
	dfs(d+1,s&se[d][0],-v);
	dfs(d+1,s&se[d][1],-v);
}
int main(){
	scanf("%d%d%d%d",&n,&K,&S,&T);
	for(i=**C=1;i<=n;++i)for(j=*C[i]=1;j<=i;++j)C[i][j]=C[i-1][j-1]+C[i-1][j];
	if((T&S)!=S){puts("0");return 0;}
	for(i=0;i<n;++i){
		scanf("%d",a+i);
		if((a[i]&S)==S && (a[i]&T)==a[i])all|=1ll<<i;
	}
	for(i=0;i<18;++i)if((S^T)>>i&1){
		for(j=0;j<n;++j)se[m][a[j]>>i&1]|=1ll<<j;
		se[m][0]&=all;se[m++][1]&=all;
	}
	dfs(0,all,1);
	for(i=1;i<=K;++i)for(j=i;j<=n;++j)ans+=b[j]*C[j][i];
	printf("%lld\n",ans);
}