Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

2 2

Sample Output 1

8

We will denote the dogs by A and B, and the monkeys by C and D. There are eight possible arrangements: ACBD, ADBC, BCAD, BDAC, CADB, CBDA, DACB and DBCA.

Sample Input 2

3 2

Sample Output 2

12

Sample Input 3

1 8

Sample Output 3

0

Sample Input 4

100000 100000

Sample Output 4

530123477

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Queue;
import java.util.Random;
import java.util.Scanner;
 


public class Main {
	private static Scanner sc = new Scanner(System.in);
	//result = Math.pow(num1, num3)
	//StringBuffer str = new StringBuffer(hoge1);
	//String hoge2 = str.reverse().toString();
	//map.containsKey(A)

	//Map<String, Integer> map = new HashMap<String, Integer>(n);
	/*for ( キーのデータ型 key : マップの名前.keySet() ) {
		データのデータ型 data = マップの名前.get( key );
		
		// keyやdataを使った処理;
	}*/
	//int i = Integer.parseInt(s);
	//Queue<String> qq=new ArrayDeque<>(); //add,poll,peek
	//Deque<String> qq=new ArrayDeque<>();//push,pop,peek
	//ArrayList<String> cc = new ArrayList<>(n);
	//Collections.sort(list);
	//Array.sort(list);   cc.contains(tmp)
	//Arrays.asList(c).contains("a")
	//list.set(1,"walk");//1 1 2 3 5
	static long mod =1000000007;
	//static 	ArrayList<Integer> cc = new ArrayList<>(100);
	
	
	
	
	public static void main(String[] args) {
		//int N=sc.nextInt();for(int i=0;i<N;i++) {}
		//int a[]=new int[N];
		//Ae2+Be2=Ce2
		long N=sc.nextInt();
		long M=sc.nextInt();
		
		if(N==M) {
			p(((Nb(N)*Nb(N))%mod*2)%mod);
			
			
			
		}else if(Math.abs(M-N)==1){
			p((Nb(M)*Nb(N))%mod);
			
			
			
		}else
			p(0);
		
		}
		static long Nb(long t) {
			long r=1;
			for(long i=t;i>0;i--) {
				r*=i;
				r%=mod;
				
			}
			return (int)r;
			
		}
		
	
	
	
	public static long lcm(long a, long b) {
        return a / gcd(a, b) * b;
	}
	
	public static long gcd(long a, long b) {
        return b == 0 ? a: gcd(b, a % b);
	}

	


	
	static void p(String ans) {System.out.println(ans);};
	static void p(int ans) {System.out.println(ans);};
	static void p() {System.out.println();};
	static void p(long ans) {System.out.println(ans);};
	static void p(double ans) {System.out.println(ans);};
	
}