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: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;

int ans = 0;
vector<vector<int>> field;
vector<bool> visited;

void dfs(int x, int remain);

int main() {
    int n, m, a, b;
    cin >> n >> m;
    vector<vector<int>> path(n);
    visited.resize(n);
    fill(visited.begin(), visited.end(), false);
    rep(i, m) {
        cin >> a >> b;
        path.at(a - 1).push_back(b - 1);
        path.at(b - 1).push_back(a - 1);
    }
    field = path;
    dfs(0, n - 1);
    cout << ans << endl;
}

void dfs(int x, int remain) {
    if(remain == 0) {
        ans++;
        return;
    }
    visited.at(x) = true;
    rep(i, field.at(x).size()) {
        if(!visited.at(field.at(x).at(i))) dfs(field.at(x).at(i), remain - 1);
    }
    visited.at(x) = 0;
    return;
}