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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>
int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    int i;
    int a[30], b[30];
    for (i = 0; i < m; i++)
        scanf("%d %d", &a[i], &b[i]);
    int p[10][10] = {};
    for (i = 0; i < m; i++)
        p[a[i] - 1][b[i] - 1] = p[b[i] - 1][a[i] - 1] = 1;
    int x, now, y;
    int j, k;
    int q[10];
    int ans = 0;
    for (j = 0; j < pow(n, n - 1); j++)
    {
        for (i = 0; i < n; i++)
            q[i] = 0;
        x = j;
        now = x % n;
        x /= n;
        k = 0;
        q[now] = 1;
        for (i = 1; i < n; i++)
        {
            y = x % n;
            x /= n;
            if (q[y] == 1)
            {
                k++;
                break;
            }
            if (p[now][y] == 0)
            {
                k++;
                break;
            }
            now = y;
            q[now] = 1;
        }
        if (k == 0)
            ans++;
    }
    printf("%d\n", ans);
    return 0;
}