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 <cstdio>
#include <vector>
#include <queue>

long long C[51][51];


int main()
{
  int n,K;
  scanf("%d%d",&n,&K);
  for(int i=0;i<=n;i++){
    C[i][0]=C[i][i]=1ll;
    for(int j=1;j<i;j++){
      C[i][j]=C[i-1][j-1]+C[i-1][j];
    }
  }
  int S,T;
  scanf("%d%d",&S,&T);
  std::vector<int> A(n);
  for(int i=0;i<n;i++){
    scanf("%d",&A[i]);
  }
  std::vector<long long> M(18,0ll);
  for(int j=0;j<18;j++){
    for(int i=0;i<n;i++){
      if((A[i]>>j)&1){
	M[j]|=(1ll<<i);
      }
    }
  }
  std::queue<std::pair<int,long long> > Q;
  long long ans=0ll;
  if((S&T)==S){
    long long BI=(1ll<<n)-1ll;
    std::vector<int> V;
    for(int j=0;j<18;j++){
      if(((S>>j)&1)==0&&((T>>j)&1)==1){
	V.push_back(j);
      }
      else if(((T>>j)&1)==0){
	BI&=~M[j];
      }
      else{
	BI&=M[j];
      }
    }
    int m=V.size();
    for(int s=0;s<n;s++){
      if((A[s]&S)!=S||(A[s]&T)!=A[s]){
	continue;
      }
      for(int J=0;J<(1<<m);J++){
	long long I=BI;
	int sg=1;
	for(int x=0;x<m;x++){
	  if((J>>x)&1){
	    sg=-sg;
	    if((A[s]>>V[x])&1){
	      I&=M[V[x]];
	    }
	    else{
	      I&=~M[V[x]];
	    }
	  }
	}
	//printf("%d %lld\n",J,I);
	int c=0;
	for(int i=s+1;i<n;i++){
	  if((I>>i)&1){
	    c++;
	  }
	}
	for(int x=0;x<K&&x<=c;x++){
	  ans+=sg*C[c][x];
	}
      }
    }
  }
  printf("%lld\n",ans);
  return 0;
}
