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 <vector>
#include <map>
using namespace std;

int main()
{
    int N, K, S, T;
    cin>>N>>K>>S>>T;
    vector<int> A(N);
    for (int &a: A)
        cin>>a;

    if ((S&T)!=S)
    {
        cout<<0<<endl;
        return 0;
    }

    int B = 0;
    for (int b=0; b<18; b++)
        if ((S>>b&1)==0 && (T>>b&1)==1)
            B++;
    
    vector<int> A_;
    A_.swap(A);
    for (int a: A_)
        if ((a&S)==S && (a|T)==T)
        {
            int t = 0;
            for (int b=0; b<18; b++)
                if ((S>>b&1)==0 && (T>>b&1)==1)
                    t = t<<1|a>>b&1;
            A.push_back(t);
        }
    N = (int)A.size();

    vector<vector<long long>> C;
    for (int i=0; i<=N; i++)
    {
        C.push_back(vector<long long>(i+1));
        C[i][0] = C[i][i] = 1;
        for (int j=1; j<i; j++)
            C[i][j] = C[i-1][j-1]+C[i-1][j];
    }

    long long ans = 0;
    map<int, int> M;
    for (int b=0; b<1<<B; b++)
    {
        M.clear();
        for (int a: A)
            M[a&b]++;
        long long a = 0;
        for (auto m: M)
            for (int k=1; k<=K; k++)
                if (m.second>=k)
                    a += C[m.second][k];
        int c = 0;
        for (int i=0; i<B; i++)
            c += b>>i&1;
        ans += c%2==0 ? a : -a;
    }
    cout<<ans<<endl;
}
