Codefoces 986C AND Graph（DFS）

C. AND Graph

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a set of size $m$ with integer elements between $0$ and $2^n-1$ inclusive. Let's build an undirected graph on these integers in the following way: connect two integers $x$ and $y$ with an edge if and only if $x\&y=0$. Here $\&$ is the bitwise AND operation. Count the number of connected components in that graph.

Input

In the first line of input there are two integers $n$ and $m$ ($0\le n\le 22$, $1\le m\le 2^n$).

In the second line there are $m$ integers $a_1,a_2,\dots ,a_m$ ($0\le a_i<2^n$) — the elements of the set. All $a_i$ are distinct.

Output

Print the number of connected components.

Examples

input

Copy

2 3
1 2 3

output

Copy

2

input

Copy

5 5
5 19 10 20 12

output

Copy

2

Note

Graph from first sample:

Graph from second sample:

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
#include<queue>
#define ll long long
using namespace std;
int f[35];
int n,m,isp[5001000],a[5001000],vis[5001000],N,ans=0,i;
void dfs(int x)
{
//	printf("%d %d\n",x,i);
	if(vis[x]==1) return ;
	vis[x]=1;
	if(isp[x]) dfs(N^x);
	for(int i=1;i<=n;i++) if(x&(1<<(i-1))) dfs(x^(1<<(i-1)));
}
void solve()
{
	for(int i=1;i<=m;i++)
	   if(!vis[a[i]])
	     {
	     	ans++;
	     	vis[a[i]]=1;
	     	dfs(N^a[i]);
		 }
}
int main()
{
f[1]=1;
for(int i=2;i<=30;i++) f[i]=f[i-1]*2;
scanf("%d%d",&n,&m);
N=f[n+1]-1;
for(int i=1;i<=m;i++) scanf("%d",&a[i]);
for(int i=1;i<=m;i++) isp[a[i]]=1;
solve();
printf("%d\n",ans);	
}