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;
using LL = long long;
constexpr int maxn = 50;
int A[maxn + 1];
LL C[maxn + 1][maxn + 1];
int cnt[1 << 18];
int main(){
    int N, K, S, T;
    cin >> N >> K >> S >> T;
    for(int i = 1; i <= N; i += 1){
        cin >> A[i];
        if((A[i] & S) != S or (A[i] | T) != T) A[i] = -1;
    }
    for(int i = 0; i <= N; i += 1)
        for(int j = 0; j <= i; j += 1)
            C[i][j] = j ? C[i - 1][j] + C[i - 1][j - 1] : 1;
    for(int i = 1; i <= N; i += 1)
        for(int j = 2; j <= N; j += 1)
            C[i][j] += C[i][j - 1];
    LL ans = 0;
    for(int i = 0; i < (1 << 18); i += 1){
        if((i & S) or (i & T) != i) continue;
        for(int j = 1; j <= N; j += 1) if(~A[j]) cnt[A[j] & i] += 1;
        LL pans = 0;
        for(int j = 1; j <= N; j += 1) if(~A[j]){
            pans += C[cnt[A[j] & i]][K];
            cnt[A[j] & i] = 0;
        }
        if(__builtin_parity(i)) ans -= pans;
        else ans += pans;
    }
    cout << ans << "\n";
    return 0;
}