C

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 109+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≤105

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of possible arrangements, modulo 109+7.

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Sample Input 1

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2 2

Sample Output 1

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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.

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Sample Input 2

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3 2

Sample Output 2

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12

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Sample Input 3

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1 8

Sample Output 3

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0

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Sample Input 4

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100000 100000

Sample Output 4

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530123477

大数据的求模；

         该题是求两种动物不相邻排序方法的个数，所以若两种动物的数量差值大于等于2，则没有这种排序方法；

         因此只有当差值 t 为1或0时可求，阶乘的模可以等于阶乘中的每个数的模的乘积再求模；

#include <iostream>#include <cstdio>#include <string>#include <cstdio>#include <cstring>#include <algorithm>#include <iomanip>#include <cmath>using namespace std;int main(){    int a,b;    while(cin>>a>>b){        if(abs(a-b)>=2)            cout<<0<<endl;        else if(a==b){            long long int sum=1;            for(int i=1;i<=a;i++){                sum*=i;                sum=sum%(1000000000+7);            }            sum=sum*sum*2%(1000000000+7);            cout<<sum<<endl;        }        else{            long long int sum=1;            long long int summ=1;            for(int i=1;i<=max(a,b);i++){                sum*=i;                sum=sum%(1000000000+7);            }            for(int i=1;i<=min(a,b);i++){                summ*=i;                summ=summ%(1000000000+7);            }            sum=sum*summ%(1000000000+7);            cout<<sum<<endl;        }    }    return 0;}