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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#define MOD 1000000007
#define MAX 510000

long int FAC[MAX],FINV[MAX],INVE[MAX];
long int SUM(long int a,long int b){return (a%MOD+b%MOD+MOD)%MOD;}//a+b
long int PROD(long int a,long int b){return a%MOD*(b%MOD)%MOD;}//a*b
long int POW(long int a,long int b){//a^b
  long int c=1;
  for(a%=MOD;b;a=PROD(a,a),b>>=1)if(b&1)c=PROD(a,c);
  return c;
}
long int INV(long int a){return a<MAX?INVE[a]:POW(a,MOD-2);}//a^-1
long int QUOT(long int a,long int b){return PROD(a,INV(b));}//a/b
void init_inv(long int a){//make inv list
  INVE[1]=1;
  for(long int i=2;i<=a;i++)INVE[i]=SUM(0,-PROD(INVE[MOD%i],MOD/i));
}
void init_com(long int a){//make fac and facinv list
  FAC[0]=FAC[1]=1;
  FINV[0]=FINV[1]=1;
  init_inv(a);
  for(long int i=2;i<=a;i++){
    FAC[i]=PROD(FAC[i-1],i);
    FINV[i]=PROD(FINV[i-1],INV(i));
  }
}
long int COM(long int a,long int b){return PROD(PROD(FAC[a],FINV[a-b]),FINV[b]);}

int main(){
  long int m,n;
  scanf("%ld%ld",&n,&m);
  if(m-n>1||m-n<-1){printf("0");return 0;}
  init_com(n>m?n:m);
  printf("%ld",PROD(PROD(FAC[n],FAC[m]),1+(n==m)));
}