枚举，用并查集来判断图的连通性，样例过不了，求调

#include<bits/stdc++.h>
using namespace std;
#define int long long

const int mod = 1e9 + 7, N = 32;

int fa[N], n, m, ans;
bool d[N], b[N];


struct edge{
	int from, to;
} e[N];


int find(int x) {
	if (fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}

void merge(int x, int y) {
	int f1 = find(x), f2 = find(y);
	if (f1 != f2) 
		fa[f2] = f1;
}
int now[N];
bool check() {
	for (int i = 1; i <= n; i++) fa[i] = i, now[i] = 0;
	int cnt = 0;
	for (int i = 1; i <= m; i++) {
		if (b[i]) {
			merge(e[i].from, e[i].to);
		}
	}
	for (int i = 1; i <= n; i++) {
		if (d[i]) {
			now[++cnt] = i;
		}
	}
	for (int i = 2; i <= cnt; i++) {
		if (find(now[i]) != find(now[i - 1]) ) return 0;
	}
	return 1;
}

void meijub(int deep) {
	if (deep > m) {
		if (check()) ans++;
		return;	
	}
	b[deep] = 0;
	meijub(deep + 1);
	b[deep] = 1;
	meijub(deep + 1);
	return;
}
int isd = 0;
void meijud(int deep) {
	if (deep > n) {
		if (isd == 0) {
			isd = 1;
			return;
		}
		meijub(1);
		return;
	}
	d[deep] = 0;
	meijud(deep + 1);
	d[deep] = 1;
	meijud(deep + 1);
	return;
}


signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		cin >> e[i].from >> e[i].to;
	}
	meijud(1);
	cout << ans;
		
	return 0;
}