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 Python (3.4.3) (judge verdict: Accepted)

# -*- coding: utf-8 -*-

import sys
import os

N, M = map(int, input().split())
# Node
# M: Edge num

# 隣接リスト
E = [[] for i in range(N)]
#print(E)

for i in range(M):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    E[a].append(b)
    E[b].append(a)
#print(E)

# BFSだとvisitedリストも増えすぎる
# DFSがよい

visited = [False] * N

one_stroke_num = 0
def dfs(v):
    #print('dfs', v)
    global one_stroke_num

    if visited.count(True) == N:
        one_stroke_num += 1
        return
    else:
        # vから行けるところあるか
        for next_i in E[v]:
            if not visited[next_i]:
                # そこに行く
                visited[next_i] = True
                dfs(next_i)
                visited[next_i] = False
        # もう行くところない
        return

visited[0] = True
dfs(0)

print(one_stroke_num)