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;
   #define int long long
const int MAX =2000007;
const int MOD = 998244353;
//

     signed main(){
         int n,k,s,t;
         cin>>n>>k>>s>>t;
         vector<int> a1(n),a2,b;
         for(int i=0;i<n;i++)cin>>a1[i];
         for(int i=17;i>=0;i--){
             if((s&(1<<i))>(t&(1<<i))){
                 cout<<0<<endl;
                 return 0;
             }
             if((s&(1<<i))<(t&(1<<i))){
                 b.push_back(i);
                 continue;
             }
             a2.clear();
             for(auto&& j:a1)if((j&(1<<i))==(s&(1<<i)))a2.push_back(((j>>(i+1))<<i)+(j%(1<<i)));
             swap(a2,a1);
         }
         int ans=0;
        int dp[a1.size()+1][k+1];
        for(int i=0;i<=a1.size();i++){
            dp[i][0]=1;
            for(int j=1;j<=k;j++)dp[i][j]=(dp[i][j-1]*(i+1-j))/j;
        }
        for(int i=0;i<=a1.size();i++)for(int j=1;j<=k;j++)dp[i][j]+=dp[i][j-1];
       
        for(int i=0;i<(1<<b.size());i++){
              int k2=1;
            for(int j=0;j<b.size();j++){
                if(i&(1<<j))k2*=(-1);
            }
           
            map<int,int>mp;
            for(auto&& j:a1)mp[j&i]++;
            for(auto&& j:mp)ans+=k2*(dp[j.second][min(k,j.second)]-1);
           
        }
        cout<<ans<<endl;

     }


 


 






    





      

        
