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 <vector>
using namespace std;
vector<int> v[9]; int N, M;
struct Node {
    int now = 1, pre = 0; int cnt[9] = {};
};
int next(Node n) {
    if(n.cnt[n.pre] > 1) return 0;
    if(n.cnt[0] == N-1) return 1;
    int root = 0;
    for(int x : v[n.now]) {
        if(x == n.pre) continue;
        Node n2 = n;
        n2.cnt[n.now] += 1; n2.cnt[0] += 1;
        n2.now = x; n2.pre = n.now;
        root += next(n2);
    }
    return root;
}
int main() {
    cin >> N >> M;
    for(int i = 0, a, b; i < N; i++) {
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    Node n;
    cout << next(n);
}