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>

void solve(int P, int V, int *num, int N, int **Path){
	int i, p[] = {2, 3, 5, 7, 11, 13, 17, 19};
	if(P == 1){
		(*num)++;
	}
	else{
		for(i = 0; i < N; i++){
			if(P%p[i] == 0 && Path[V][i] == 1){
				solve(P/p[i], i, num, N, Path);
			}
		}
	}
}

int main(){
	int N, M, a, b, i, j, P = 1, p[] = {2, 3, 5, 7, 11, 13, 17, 19};
	scanf("%d%d", &N, &M);
	for(i = 0; i < N; i++){
		P *= p[i];
	}
	int **Path;
	Path = (int **)malloc(sizeof(int *) * N);
	for(i = 0; i < N; i++){
		Path[i] = (int *)malloc(sizeof(int) * N);
		for(j = 0; j < N; j++){
			Path[i][j] = -1;
		}
	}
	for(i = 0; i < M; i++){
		scanf("%d%d", &a, &b);
		Path[a-1][b-1] = 1;
		Path[b-1][a-1] = 1;
	}
	int *num;
	num = (int *)malloc(sizeof(int));
	*num = 0;
	solve(P/2, 0, num, N, Path);
	printf("%d\n", *num);
	free(num);
	for(i = 0; i < N; i++){
		free(Path[i]);
	}
	free(Path);
	return 0;
}