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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
	int N, M;
	cin >> N >> M;

	vector<int> order(N);
	for (int i = 0; i < N; i++) order[i] = i + 1;

	vector<pair<int, int>> ab(N);
	for (int i = 0; i < M; i++) {
		cin >> ab[i].first;
		cin >> ab[i].second;
	}

	int count = 0;

	do {
		bool ok = false;
		if (order[0] != 1) continue;
		for (int i = 1; i < N; i++) {
			bool exist = false;
			for (int j = 0; j < M; j++) {
				if (order[i - 1] == ab[j].first && order[i] == ab[j].second) exist = true;
				if (order[i - 1] == ab[j].second && order[i] == ab[j].first) exist = true;
				if (exist) {
					ok = true;
					break;
				}
			}
			if (!ok) break;
		}
		if (ok) count++;
	} while (next_permutation(order.begin(), order.end()));

	cout << count << endl;

	return 0;
}