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;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 50;
ll ch[N + 1][N + 1];

int bc(int x){
	int an = 0;
	while (x){
		++an;
		x ^= x & -x;
	}
	return an;
}

ll go(vector<int> v, int b, int k, int x){
	int n = v.size();
	f(i, 0, n)v[i] ^= x;
	ll an = 0;
	f(i, 0, 1 << b){
		int s = 0;
		f(j, 0, n)if (!(v[j] & i))++s;
		if (bc(i) & 1)an -= ch[s][k];
		else an += ch[s][k];
	}
	return an;
}

ll solve(vector<int> const &v, int b, int k){
	ll an = 0;
	f(i, 0, v.size())an += go(vector<int>(v.begin() + i + 1, v.end()), b, k - 1, v[i]);
	return an;
}

int main(){
	int n, k, s, t;
	scanf("%d%d%d%d", &n, &k, &s, &t);
	if ((s & t) != s){
		printf("0\n");
		return 0;
	}
	
	f(i, 0, n + 1){
		ch[i][0] = 1;
		f(j, 1, i + 1)ch[i][j] = ch[i - 1][j] + ch[i - 1][j - 1];
	}
	f(i, 0, n + 1)f(j, 1, n + 1)ch[i][j] += ch[i][j - 1];
	
	vector<int> v;
	int z = s ^ t, b = 0;
	f(i, 0, 18)if (z >> i & 1)++b;
	f(i, 0, n){
		int a;
		scanf("%d", &a);
		if ((a & s) != s || (a & t) != a)continue;
		int j = 0, an = 0;
		f(k, 0, 18)if (z >> k & 1)an |= (a >> k & 1) << j++;
		v.push_back(an);
	}
	printf("%lld\n", solve(v, b, k));
}