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: Wrong Answer)



import java.io.*;
import java.lang.*;
import java.util.*;
import java.math.*;

class Solver extends Main{
	Solver(){
		
	}
}//Solver_end

@SuppressWarnings("unchecked")
class Main{
	static int imax=Integer.MAX_VALUE;
	static long lmax=Long.MAX_VALUE;
	
	public static void main(String[] args) throws FileNotFoundException{
		
		Scanner sc=new Scanner(System.in);
		
		int[] num=readlnI(sc," ");
		int n=num[0], m=num[1];
		puts(n,m);
		long ns=1, ms=1;
		for(int i=1; i<=n; i++){
			ns*=i;
		}
		
		for(int i=1; i<=m; i++){
			ms*=i;
		}
		
		puts(ns,ms);
		
		if(n==m){
			puts((ns*ms)*2);
		}else if(Math.abs(m-n)==1){
			puts(ns*ms);
		}else if(Math.abs(m-n)>=2){
			puts(0);
		}
		
	}//main_end
	
//---------------------------------------------------------------------------
	static int readI(Scanner sc){
		return Integer.parseInt(sc.nextLine().trim());
	}
	static long readL(Scanner sc){
		return Long.parseLong(sc.nextLine().trim());
	}
	static String readS(Scanner sc){
		return sc.nextLine().trim();
	}
	static int[] readlnI(Scanner sc, String dl){
		return Arrays.stream(sc.nextLine().trim().split(dl)).mapToInt(Integer::parseInt).toArray();
	}
	static long[] readlnL(Scanner sc, String dl){
		return Arrays.stream(sc.nextLine().trim().split(dl)).mapToLong(Long::parseLong).toArray();
	}
	static String[] readlnS(Scanner sc, String dl){
		return sc.nextLine().trim().split(dl);
	}
	@SuppressWarnings("unchecked")
	static <T> void puts(T... args){
		for(int i=0; i<args.length-1; i++){
			System.out.printf("%s ",args[i]);
		}
		System.out.println(args[args.length-1]);
	}
	static void puts(int[] ary){
		for(int i=0; i<ary.length-1; i++){
			System.out.printf("%d ",ary[i]);
		}
		System.out.println(ary[ary.length-1]);
	}
	static void puts(long[] ary){
		for(int i=0; i<ary.length-1; i++){
			System.out.printf("%d ",ary[i]);
		}
		System.out.println(ary[ary.length-1]);
	}
	static void bs(){
		System.out.println("");
	}
	static int sum(int[] ary){
		int sum=0;
		for(int i=0; i<ary.length; i++) sum+=ary[i];
		return sum;
	}
	static long sum(long[] ary){
		long sum=0;
		for(int i=0; i<ary.length; i++) sum+=ary[i];
		return sum;
	}
	static void tblSort(int[][] tbl){
		Arrays.sort(tbl, new Comparator<int[]>(){
			@Override
			public int compare(int[] a, int[] b){
				if(a[0]==b[0]){
					return b[1]-a[1];
				}else{
					return b[0]-a[0];
				}
			}
		});
	}
}