Problem: 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.

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Figure 1: an example of an undirected graph

The following path shown in Figure 2 satisfies the condition.

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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.

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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.

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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.

Examples

Input

3 3
1 2
1 3
2 3


Output

2


Input

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7


Output

1

Solution (in PYTHON):

Solution:
import copy
s=list()

N,M=map(int,raw_input().split())
n=[[0 for i in range(N)] for j in range(N)]
for i in range(M):
        a,b=map(int,raw_input().split())
        n[a-1][b-1]=1
        n[b-1][a-1]=1
arrived=[0 for i in range(N)]

arrived[0]=1
def solve(s,arrived):
        narrived=copy.deepcopy(arrived)
        global n
        global N
        op=False
        ans=0
        for i in range(N):
                arrived=copy.deepcopy(narrived)
                if(s[i]==1 and arrived[i]!=1):
                        op=True
                        arrived[i]=1
                        ans+=solve(n[i],arrived)
        
        if(sum(arrived)==N and op==False):
                return 1
        elif(op==False):
                return 0
        return ans
                        
print(solve(n[0],arrived))