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 <string.h>
// #define INF 1001001000
// #define NO_PRINT 70
// int main() {
// 	int n, m, a[50], b[50], s[50], numCnt[50] = {0};
// 	for (int i = 0; i < 50; i++) {
// 		s[i] = INF;
// 	}
// 	char cmd[15];
// 	scanf("%d", &n);
// 	for (int i = 0; i < n; i++) {
// 		scanf("%d", &a[i]);
// 	}
// 	scanf("%d", &m);
// 	for (int i = 0; i < m; i++) {
// 		scanf("%d", &b[i]);
// 	}
// 	scanf("%s", cmd);
// 	if (strcmp(cmd, "intersection") == 0) {
// 	} else if (strcmp(cmd, "union_set") == 0) {
// 	} else if (strcmp(cmd, "symmetric_diff") == 0) {
// 	} else if (strcmp(cmd, "subtract") == 0) {
// 		int x;
// 		scanf("%d", &x);
// 		for (int i = 0; i < n; i++) {
// 			if (a[i] == x) {
// 				s[i] = NO_PRINT;
// 			} else {
// 				s[i] = a[i];
// 			}
// 		}
// 	} else if (strcmp(cmd, "increment") == 0) {
// 		for (int i = 0; i < n; i++) {
// 			s[i] = (a[i] + 1) % 50;
// 		}
// 	} else if (strcmp(cmd, "decrement") == 0) {
// 		for (int i = 0; i < n; i++) {
// 			s[i] = a[i] - 1;
// 			if (s[i] == -1)
// 				s[i] = 49;
// 		}
// 	}
// 	for (int i = 0; s[i] != INF; i++) {
// 		if (s[i] != NO_PRINT) {
// 			printf("%d ", s[i]);
// 		}
// 	}
// 	return 0;
// }

#include <stdbool.h>
#include <stdio.h>
int a[8], ans = 0;
bool G[8][8] = {0}, isUsed[8] = {0};
void perm(int n, int len) {
	if (len == n) {
		if (a[0] != 0) {
			return;
		}
		bool isOK = true;
		for (int i = 1; i < n; i++) {
			// printf("%d ", a[i]);
			if (!G[a[i]][a[i - 1]]) {
				isOK = false;
				break;
			}
		}
		// printf("\n");
		if (isOK)
			ans++;
		return;
	}
	for (int i = 0; i < n; i++) {
		if (!isUsed[i]) {
			a[len] = i;
			isUsed[i] = true;
			perm(n, len + 1);
			isUsed[i] = false;
		}
	}
}
int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		a--;
		b--;
		G[a][b] = true;
		G[b][a] = true;
	}
	perm(n, 0);
	printf("%d\n", ans);
	return 0;
}