Problem:
Score : 300 points

Problem Statement

You are given an undirected unweighted graph with N vertices and M edges that contains neither self-loops nor double edges.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

How many different paths start from vertex 1 and visit all the vertices exactly once?

Here, the endpoints of a path are considered visited.

For example, let us assume that the following undirected graph shown in Figure 1 is given.

Figure 1: an example of an undirected graph

The following path shown in Figure 2 satisfies the condition.

Figure 2: an example of a path that satisfies the condition

However, the following path shown in Figure 3 does not satisfy the condition, because it does not visit all the vertices.

Figure 3: an example of a path that does not satisfy the condition

Neither the following path shown in Figure 4, because it does not start from vertex 1.

Figure 4: another example of a path that does not satisfy the condition

Constraints

2≦N≦8

0≦M≦N(N-1)/2

1≦a_i<b_i≦N

The given graph contains neither self-loops nor double edges.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the number of the different paths that start from vertex 1 and visit all the vertices exactly once.

Sample Input 1

3 3
1 2
1 3
2 3

Sample Output 1

2

The given graph is shown in the following figure:

The following two paths satisfy the condition:

Sample Input 2

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7

Sample Output 2

1

This test case is the same as the one described in the problem statement.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <iomanip>
#include <bitset>
#include <set>
#include <map>
#include <stdio.h>
#include <numeric>
#include <cstring>

#define rep(i,n) for (int i = 0; i < (n); i++)
#define rep1(i,n) for (int i = 1; i < (n); i++)
#define FOR(i,a,b) for (int i=(a); i < (b); i++)
#define MOD 1000000007 //10^9+7
using namespace std;
using ll = long long;
using PII = pair<int, int>;
const int INF = numeric_limits<int>::max();

// 負の数にも対応した % 演算
long long mod(long long val, long long m) {
    long long res = val % m;
    if (res < 0) res += m;
    return res;
}

//greatest common divisor
long long gcd(ll a, ll b) 
{
    if (a % b == 0) {
        return b;
    } else {
        return gcd(b, a % b);
    }
}

//least common multiple
long long lcm(ll a, ll b)
{
    return a / gcd(a, b) * b ;
}

int main()
{
    int n, m; cin >> n >> m;
    vector<vector<int>> a(n);
    rep(i,m) {
        int ai, bi;
        cin >> ai >> bi;
        ai--; bi--;
        a[ai].push_back(bi);
        a[bi].push_back(ai);
    }

    int path[n];
    rep(i,n) path[i] = i;

    int ans = 0;
    do {
        int reach = 1;
        rep(i,n-1) {
            int flag = 0;
            rep(j,a[i].size()) {
                if(a[i][j] == path[i+1]) {
                    flag = 1;
                }
            }
            if (flag == 0) {
                reach = 0;
                break;
            }
        }
        if (reach)
            ans++;
    } while(next_permutation(path+1, path+n));

    cout << ans++ << endl;
                
    return 0;
}