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 <bits/stdc++.h>
using namespace std;

int n, m;
vector<int> a, b;
vector<int> flags(10);
vector<vector<int>> table(10, vector<int>(10));

int recurse(int num) {

    int flag = 1;
    for (int i = 1; i <= n && flag; i++) flag = flags[i];
    if (flag) return 1;
 
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        if (table[num][i] && !flags[i]) {
            flags[i] = 1;
            sum += recurse(i);
            flags[i] = 0;
        }
    }

    return sum;
}

int main() {
    cin >> n >> m;
    a.resize(m); b.resize(m);
    for(int i = 0; i < m; i++) cin >> a[i] >> b[i];
    for(int i = 0; i < m; i++) table[a[i]][b[i]] = table[b[i]][a[i]] = 1;
    flags[1] = 1;
    int ans = recurse(1);
    cout << ans << endl;
    return 0;
}