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 JAVA):

Solution:
import java.util.*;

class Main {
	static int cnt;
	static int[] flag;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int[] a = new int[M];
		int[] b = new int[M];
		flag = new int[N];
		cnt = 0;
		for (int i=0;i<M;i++) {
			a[i]=sc.nextInt();
			b[i]=sc.nextInt();
		}
		int[][] next = new int[N][N];
		for (int i=0;i<N;i++) {
			for (int j=0;j<N;j++) {
				for (int k=0;k<M;k++) {
					if (a[k]-1==i && b[k]-1==j) {
						next[i][j]=1;
					}
					if (a[k]-1==j && b[k]-1==i) {
						next[i][j]=1;
					}
				}
			}
		}
		flag[0]=1;
		DFS(0,flag,N,M,next);
		System.out.println(cnt);
	}

	static void DFS(int v,int[] flag,int N,int M,int[][] next) {
		boolean bool=false;
		for (int i=0;i<N;i++) {
			if (flag[i]==0) {
				bool = false;
				break;
			}
			if (i==N-1) {
				bool = true;
			}
		}
		if (bool) {
			cnt++;
			return;
		}
		for (int i=0;i<N;i++) {
			if (next[v][i]==1 && flag[i]==0) {
				flag[i]=1;
				DFS(i,flag,N,M,next);
				flag[i]=0;
			} else {
				continue;
			}
		}
		return;
	}
}