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 <stdlib.h>
#include <string.h>
#include <math.h>
#define rep(i, n) for (i = 0; i < n; i++)
#define rrep(i, n) for (i = n; i >= 0; i--)
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)

int n, m, g[9][9], v[9] = { 0, 1 }, ans;

void dfs(int i, int s) {
    int k;
    for (k = 1; k <= n; k++) {
        if (g[i][k] == 1 && v[k] == 0) {
            v[k] = 1;
            dfs(k, s + k);
        }
    }
    v[i] = 0;
    if (s == n * (n + 1) / 2) ans++;
    return;
}

int main() {
    scanf("%d %d", &n, &m);

    int a, b, i;
    rep(i, m) {
        scanf("%d %d", &a, &b);
        g[a][b] = g[b][a] = 1;
    }

    dfs(1, 1);
    printf("%d\n", ans);
    return 0;
}