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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using static System.Console;
using static System.Math;
 
public class Solve{
    static public int mod = 1000000007;
    public static void Main(){
        // 方針
        // 
        int[] n = inta();
        if(Abs(n[1]-n[0]) >1){
            WriteLine(0);
        }else{
            long ans = 1;
            for(int i=n[0];i>1;i--){
                ans = ans*i%mod;
            }
            for(int i=n[1];i>1;i--){
                ans = ans*i%mod;
            }
            if(n[0] == n[1]){
                ans = ans*2%mod;
            }
            WriteLine(ans);
        }
        
        
        
        
        
    }
    static int binarysearchd(int[] a,int b){
        int leng = a.Length;
        int min = 0;
        int max = leng-1;
        while(max - min > 1){
            int mid = (min+max)/2;
            if(a[mid] < b){
                min = mid;
            }else{
                max = mid;
            }
        }
        if(a[0] >= b){
            return 0;
        }else if(a[leng-1] < b){
            return leng;
        }else{
            if(a[max] < b){
                return max+1;
            }
            return min+1;
        }
    }
    static int binarysearchu(int[] a,int b){
        int leng = a.Length;
        int min = 0;
        int max = leng-1;
        while(max - min > 1){
            int mid = (min+max)/2;
            if(a[mid] < b){
                min = mid;
            }else{
                max = mid;
            }
        }
        if(a[0] > b){
            return leng;
        }else if(a[leng-1] <= b){
            return 0;
        }else{
            if(a[max] <= b){
                return leng-(max+1);
            }
            return leng-(min+1);
        }
    }
    // 1からaまでの数字並べ替え全通り返すやつ
    static List<string> permutation(int a){var list = new List<string>();if(a == 1){list.Add("1");return list;}else{foreach(var k in permutation(a-1)){var ch = k.ToCharArray();for(int i=0;i<=ch.Length;i++){string intemp = "";if(i == 0){intemp += a.ToString();}for(int j=0;j<ch.Length;j++){intemp += ch[j].ToString();if(i == j+1){intemp += a.ToString();}}list.Add(intemp);}}}return list;}
    public static void swap(ref int a,ref int b){int temp = a;a= b;b = temp;}
    static void charswap(ref char a,ref char b){char temp = a;a= b;b = temp;}
    static int ncr(int n,int r){r = Min(r,n-r);long nn = 1;for(int i=n-r+1;i<=n;i++){nn = nn*i%mod;}long rr = 1;for(int i=1;i<=r;i++){rr = rr*i%mod;}rr = square((int)rr,mod-2);nn = nn * rr %mod;return (int)nn;}
    // a^b mod
    static int square(int a,int b){string binary = Convert.ToString(b,2);int bileng = binary.Length;long a_power = a;long value = 1;for(int i=bileng-1;i>=0;i--){if(binary[i] == '1'){value = value*a_power%mod;}a_power = a_power*a_power%mod;}return (int)value;}
    static int square2(int a,int b){long output = 1;var list = new List<long>();int sh = 1;long n = a;list.Add(a);while(sh < b){sh *= 2;n = n*n%mod;list.Add(n);}for(int i=list.Count-1;i>=0;i--){if(b > sh){b -= sh;sh /= 2;output = output*list[i]%mod;}}return (int)output;}
    //各種読取
    static string rstr(){ return ReadLine(); }
    static int rint(){ return int.Parse(ReadLine()); }
    static long rlong(){ return long.Parse(ReadLine()); }
    static string[] stra(){ return ReadLine().Split(' '); }
    static char[] chara(){ string[] a=stra();string b="";for(int i=0;i<a.Length;i++){b+=a[i];}return b.ToCharArray();}
    static int[] inta(){ string[] read_str_array = ReadLine().Split(' '); int[] return_int_array = new int[read_str_array.Length]; for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_int_array[countup_i] = int.Parse(read_str_array[countup_i]); } return return_int_array; }
    static int[,] inta2(int num_array,int in_array){ int[,] int_array2 = new int[num_array,in_array]; for(int i=0;i<num_array;i++){ int[] temp_array = inta(); for(int j=0;j<in_array;j++){ int_array2[i,j] = temp_array[j]; } } return int_array2; }
    static long[] longa(){ string[] read_str_array = ReadLine().Split(' '); long[] return_long_array = new long[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_long_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_long_array; }
    static double[] doublea(){ string[] read_str_array = ReadLine().Split(' '); double[] return_double_array = new double[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_double_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_double_array; }
    // -----------------------------
    static long divideup(long divided,long divid){ long temp_divide = divided/divid; if(divided % divid > 0){ temp_divide++; } return temp_divide; }
    static long GCD(long a,long b){ if(a < b){ long temp = a; a = b; b = temp; } if(a % b == 0){ return b; } else{ long temp = b; b = a%b; a = temp; return GCD(a,b); } }
    static long LCM(long a,long b){ return a * b / GCD(a,b); }
}