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: Accepted)

#include <iostream>
#include <vector>
#include <numeric>
#include <functional>

using namespace std;

void DFS(int start, const vector<vector<int>>& g, int& count, vector<bool>& visited){
    visited[start] = true;
    if (accumulate(visited.begin(), visited.end(), 1, multiplies<int>())){
        count += 1;
    }
    else{
        for (int i = 0; i < g.size(); ++i){
            if (g[start][i] == 1 && !visited[i]){
                visited[i] = true;
                DFS(i, g, count, visited);
                visited[i] = false;
            }   
        }
    }

    return;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N, M, a, b;
    cin >> N >> M;
    
    vector<vector<int>> g(N, vector<int>(N, 0));
    for (int i = 0; i < M; ++i){
        cin >> a >> b;
        g[a-1][b-1] = 1;
        g[b-1][a-1] = 1;
    }

    int count = 0;
    vector<bool> visited(N, false);

    DFS(0, g, count, visited);

    cout << count << "\n";

    return 0;
}