C. AND Graph+dfs

每次dfs尝试所有可能相连的状态

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 <bits/stdc++.h>


typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

int msk;
int n, m;
int fl[1 << 22];
int was[1 << 22];

void dfs1(int v, int c) {
	if (c == 0) {
		fl[v] = 0;
		if (!was[msk - v])
			dfs1(msk - v, 1);
	}
	else {
		was[v] = 1;
		if (fl[v])
			dfs1(v, 0);
		for (int i = 0; i < n; ++i) {
			int x = v & (msk - (1 << i));
			if (!was[x])
				dfs1(x, 1);
		}
	}
}

int main() {
    freopen("in.txt","r",stdin);
	cin >> n >> m;
	msk = (1 << n) - 1;
	for (int i = 0; i < m; ++i) {
		int x;
		cin >> x;
		fl[x] = 1;
	}
	int ans = 0;
	for (int i = 0; i < (1 << 22); ++i) {
		if (fl[i]) {
			dfs1(i, 0), ++ans;
		}
	}
	cout << ans << "\n";
	return 0;
}