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;
int n,k,s,t,a[55],b[55],cnt,ct,cc,ka[333333],kb[333333],kc[333333],pop[333333],rt;
long long c[55][55],fuck[55],ans;
void FWT_or(int a[],int flg)
{
	for (int i=2;i<=(1<<rt);i<<=1)
	{
		for (int p=i>>1,j=0;j<(1<<rt);j+=i)
		{
			for (int k=j;k<j+p;k++)
			{
				a[k+p]+=a[k]*flg;
			}
		}
	}
}
int main()
{
	scanf("%d%d",&n,&k);
	scanf("%d%d",&s,&t);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	c[0][0]=1;
	for (int i=1;i<=n;i++)
	{
		c[i][0]=1;
		for (int j=1;j<=i;j++)
		{
			c[i][j]=c[i-1][j]+c[i-1][j-1];
		}
	}
	for (int i=0;i<18;i++)
	{
		if (s&(1<<i)) continue;
		if (!(t&(1<<i))) continue;
		ct++;
	}
	for (int i=1;i<=n;i++)
	{
		if ((a[i]&s)!=s) continue;
		if (a[i]&((1<<18)-1-t)) continue;
		++cnt;cc=0;
		for (int j=0;j<18;j++)
		{
			if (s&(1<<j)) continue;
			if (!(t&(1<<j))) continue;
			if (a[i]&(1<<j)) b[cnt]|=(1<<cc);
			cc++;
		} 
	}
	for (int i=0;i<=n;i++)
	{
		for (int j=1;j<=k;j++)
		{
			fuck[i]+=c[i][j];
		}
	}
	for (int i=0;i<(1<<ct);i++) pop[i]=__builtin_popcount(i);
	for (int i=0;i<(1<<ct);i++)
	{
		rt=0;
		for (int j=0;j<ct;j++)
		{
			if (i&(1<<j)) continue;
			rt++;
		}
		for (int j=0;j<(1<<rt);j++) ka[j]=0;
		for (int j=1;j<=cnt;j++)
		{
			if ((b[j]&i)!=i) continue;
			int cmsk=0;cc=0;
			for (int k=0;k<ct;k++)
			{
				if (i&(1<<k)) continue;
				if (b[j]&(1<<k)) cmsk|=(1<<cc);
				cc++;
			}
			ka[cmsk]++;
		}
		FWT_or(ka,1);
		for (int j=0;j<(1<<rt);j++)
		{
			if ((pop[i]+rt-pop[j])&1) ans-=fuck[ka[j]];
			else ans+=fuck[ka[j]];
		}
	}
	printf("%lld\n",ans);
	return 0;
}