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>

int N, M;

int visited[10];
int connected[10][10];

int ans;

void dfs(int now,int depth);

int main() {


	scanf("%d %d",&N,&M);

	int i;

	for (i = 0; i < M; i++) {
		int a, b;
		scanf("%d %d",&a,&b);
		connected[a][b] = 1;
		connected[b][a] = 1;
	}
	dfs(1, 1);
	printf("%d\n",ans);

	return 0;
}
void dfs(int now,int depth) {

	if (depth >= N) { ans++; return; }

	visited[now] = 1;

	int i;

	for (i = 1; i <= N; i++) {
		if (connected[now][i] == 1 && visited[i] == 0) {
			visited[i] = 1;
			dfs(i,depth+1);
			visited[i] = 0;
		}
	}

}