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 <string>
#include <tuple>
#include <algorithm>
#include <numeric>
#include <climits>
#include <functional>

using namespace std;

int main()
{
    int n, m;
    cin >> n >> m;

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

    vector<bool> v(n);
    long count = 0;
    function<void(int)> iter = [&](int node){
        v[node] = true;
        bool deadend = true;
        for (int i = 0; i < m; i++) {
            if (a[i] == node && !v[b[i]]) {
                iter(b[i]);
                deadend = false;
            } else if (b[i] == node && !v[a[i]]) {
                iter(a[i]);
                deadend = false;
            }
        }
        if (deadend) {
            if (all_of(v.begin(),
                       v.end(),
                       [](bool b){return b;})) {
                count += 1;
            }
        }
        v[node] = false;
    };
    iter(0);

    cout << count << endl;

    return 0;
}