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>
#define endl "\n"
using namespace std;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

const long long INF = 1e18;
//const ll mod = 1000000007;
ll N, K, S, T;
vector<ll> A;
ll ans = 0;
ll combination[60][60];
ll combinationsum[60][60];
int main() {
    for(int i = 0; i <= 55; i++) {
        combination[i][0] = 1;
        for(int j = 1; j <= i; j++) {
            combination[i][j] = combination[i-1][j-1] + combination[i-1][j];
        }
        for(int j = 0; j <= 55; j++) {
            combinationsum[i][j+1] = combinationsum[i][j] + combination[i][j];
        }
    }
    cin >> N >> K >> S >> T;
    if((T & S) != S) {
        cout << 0 << endl;
        return 0;
    }
    for(int i = 0; i < N; i++) {
        ll a;
        cin >> a;
        if((a & S) != S) {
            i--;
            N--;
            continue;
        }
        if((a | T) != T) {
            i--;
            N--;
            continue;
        }
        //cerr << a << endl;
        A.push_back(a);
    }
    for(ll bits = 0; bits < (1LL << 18); bits++) {
        if((bits & S) != S) continue;
        if((bits | T) != T) continue;
        ll factor = __builtin_popcountll(bits ^ S);
        map<ll, ll> mp;
        for(int i = 0; i < N; i++) {
            ll a = A[i];
            a &= (~bits);
            mp[a]++;
        }
        ll num = 0;
        for(auto tmp : mp) {
            ll val = combinationsum[tmp.second][K+1] - 1;
            //cerr << bits << " " << tmp.first << " " << tmp.second << " " << val << endl;
            //num += (1LL << tmp.second);
            num += val;
        }
        if(factor % 2 == 1) ans -= num;
        else ans += num;
    }
    ans = abs(ans);
    cout << ans << endl;
    return 0;
}