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.Scanner;
import java.util.Arrays;
public class Main {
	public static void main(String[] args) {
		ABC054CSolve solve = new ABC054CSolve();
		solve.main();
	}
}

class ABC054CSolve {
	
	int N;
	int M;
	int[][] adj;
	
	int[] p;
	int count = 0;
	
	ABC054CSolve() {
		Scanner cin = new Scanner(System.in);
		this.N = cin.nextInt();
		this.M = cin.nextInt();
		this.adj = new int[N][N];
		for (int i = 0; i < M; i++) {
			int a = cin.nextInt() - 1;
			int b = cin.nextInt() - 1;
			adj[a][b] = 1;
			adj[b][a] = 1;
		}
	}
	
	void main() {
		this.p = new int[N];
		for (int i = 0; i < N; i++) { p[i] = i; }
		dfs(0, (1 << N) - 1);
		System.out.println(count);
	}
	
	void dfs(int pos, int mask) {
		if (pos == N) {
			if (isExistPath()) {
				count++;
			}
			return;
		} 
		for (int i = 0; i < N; i++) {
			if ((mask & 1 << i) >> i == 1) {
				p[pos] = i;
				dfs(pos + 1, (mask ^ (1 << i)));
			}
		}
	}
	
	boolean isExistPath() {
		if (p[0] != 0) return false;
		int curr = p[0];
		for (int i = 1; i < N; i++) {
			int next = p[i];
			if (adj[curr][next] != 1) {
				return false;
			}
			curr = next;
		}
		return true;
	}
}