C. AND Graph+dfs

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本文介绍了一个使用深度优先搜索（DFS）算法解决特定图论问题的方法：计算一个由整数通过位与运算（AND）定义的无向图中的连通组件数量。输入包含一组整数，通过位与运算确定节点间的连接，并统计连通区域。

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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 \leq n \leq 22

1 \leq m \leq 2^{n}

In the second line there are $m$

integers

a_{1}, a_{2}, \dots, a_{m}

(

0 \leq 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

Input

Copy

5 5
5 19 10 20 12

Output

Copy

Note

Graph from first sample:

$\text{[math]}$

Graph from second sample:

$\text{[math]}$

#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;
}