Snuke has N dogs and M monkeys. He wants them to line up in a row.

As a Japanese saying goes, these dogs and monkeys are on bad terms. ("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.) Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys.

How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.

Constraints

* 1 ≤ N,M ≤ 10^5

Input

Input is given from Standard Input in the following format:


N M


Output

Print the number of possible arrangements, modulo 10^9+7.

Examples

Input

2 2


Output

8


Input

3 2


Output

12


Input

1 8


Output

0


Input

100000 100000


Output

530123477
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {
	
	static final int mod = (int)1e9 + 7;
	
	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int[] fac = new int[100001];
		fac[0] = 1;
		for(int i = 1; i <= 100000; ++i)
			fac[i] = (int)(1l * fac[i - 1] * i % mod);
		int N = sc.nextInt(), M = sc.nextInt();
		if(Math.abs(N - M) > 1)
			out.println(0);
		else
		{
			long ans = (int)(1l * fac[N] * fac[M] % mod);
			if(N == M)
				ans = ans * 2 % mod;
			out.println(ans);
 		}
		
		out.close();
	}
		
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public Scanner(String s) throws FileNotFoundException{	br = new BufferedReader(new FileReader(s));}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {return Integer.parseInt(next());}

		public long nextLong() throws IOException {return Long.parseLong(next());}

		public String nextLine() throws IOException {return br.readLine();}

		public double nextDouble() throws IOException { return Double.parseDouble(next()); }

		public boolean ready() throws IOException {return br.ready();} 
	}
}