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:
#pragma GCC optimize("Ofast",3,"inline")
#include<bits/stdc++.h>
using namespace std;
#define fir first
#define sec second
#define m_p make_pair
#define y1 ygftgfgcdtfgxffgx
#define y2 yfdsesgvtyghftfvv
#define x1 xvyr6cf6fgcfgf676
#define x2 xcr6rfc5r66y6r6fr
#define up_bound upper_bound
#define low_bound lower_bound
#define next_per next_permutation
#define pb push_back
#define i_to_s to_string
typedef priority_queue<int> p_queue;
typedef priority_queue<int, vector<int>, greater<int> > min_p_queue;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
int mon[20]={31,28,31,30,31,30,31,31,30,31,30,31};
ll gcd(ll x,ll y){return ((y==0)?x:gcd(y,x%y));}
inline int read(){
	int x=0;char ch=getchar();
	while(ch<'0'||ch>'9')ch=getchar();
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x;
}

int x, c = 100000;
int n, k, s, t, a[55];
long long C[55][55], ps[55][55];
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	scanf("%d%d%d%d", &n, &k, &s, &t);
	if ((s & t) != s || (s | t) != t) {
		puts("0");
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		if ((a[i] & s) != s || (a[i] | t) != t) {
			i--;
			n--;
		}
	}
	k = min(k, n);
	for (int i = C[0][0] = 1; i <= n; i++) {
		for (int j = C[i][0] = 1; j <= i; j++)
			C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
		ps[i][1] = C[i][1];
		for (int j = 2; j <= i; j++) {
			ps[i][j] = ps[i][j - 1] + C[i][j];
		}
	}
	int fa = s ^ t;
	long long ans = 0;
	for (int mask = fa; ; mask = (mask - 1) & fa) {
		map<int, int> mp;
		for (int i = 1; i <= n; i++) {
			mp[a[i] & mask]++;
		}
		long long cur = 0;
		for (auto &it : mp) {
			cur += ps[it.second][min(k, it.second)];
		}
		if (__builtin_popcount(mask) & 1) ans -= cur;
		else ans += cur;
		if (!mask) break;
	}
	printf("%lld\n", ans);
	return 0;
}