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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define REP(i, n) for (int (i) = 0 ; (i) < (int)(n) ; ++(i))
#define REPN(i, m, n) for (int (i) = m ; (i) < (int)(n) ; ++(i))
#define REP_REV(i, n) for (int (i) = (int)(n) - 1 ; (i) >= 0 ; --(i))
#define REPN_REV(i, m, n) for (int (i) = (int)(n) - 1 ; (i) >= m ; --(i))

#define INF 2e9
#define INF_LL 1LL<<60
#define ll long long
#define MAX_V
const double PI=3.14159265358979323846;
#define MODI 1e9+7

#define print2D(h, w, arr) REP(i, h) { REP(j, w) cout << arr[i][j] << " "; cout << endl; }
#define print_line(vec, n) {for(int i=0;i<(n-1);i++) cout << (vec)[i] << " "; cout << (vec)[(n)-1] << endl;}
template<class T> void print(const T& x){cout << x << endl;}
template<class T, class... A> void print(const T& first, const A&... rest) { cout << first << " "; print(rest...); }
struct PreMain {PreMain(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(20);}} premain; 

typedef pair<ll, ll> P;
struct modint{
    using i64=int_fast64_t;
    i64 a;
    static constexpr i64 MOD=1000000007;
    modint(){a=0;}
    modint(i64 a_){
        a=a_%MOD;
        if(a<0) a+=MOD;
    }
    modint inv()const{
        i64 n=1,m=MOD-2,A=a;
        while(m){
            if(m&1)(n*=A)%=MOD;
            (A*=A)%=MOD;
            m>>=1;
        }
        modint y(n);
        return y;
    }
    bool operator==(const modint& x){
        return a==x.a;
    }
    bool operator!=(const modint& x){
        return a!=x.a;
    }
    modint& operator=(const modint& x){
        a=x.a;
        return *this;
    }
    modint operator+(const modint& x){
        modint y;
        y.a=a+x.a;
        if(y.a>MOD) y.a-=MOD;
        return y;
    }
    modint operator-(const modint& x){
        modint y;
        y.a=a-x.a;
        if(y.a<0) y.a+=MOD;
        return y;
    }
    modint operator*(const modint& x){
        modint y;
        y.a=(a*x.a)%MOD;
        return y;
    }
    modint operator/(const modint& x){
        modint y;
        y.a=(a*x.inv().a)%MOD;
        return y;
    }
    modint& operator+=(const modint& x){
        a+=x.a;
        if(a>=MOD) a-=MOD;
        return *this;
    }
    modint& operator-=(const modint& x){
        a-=x.a;
        if(a<0) a+=MOD;
        return *this;
    }
    modint& operator*=(const modint& x){
        (a*=x.a)%=MOD;
        return *this;
    }
    modint& operator/=(const modint& x){
        (a*=x.inv().a)%=MOD;
        return *this;
    }
};
istream& operator>>(istream &in,modint& x){
    int_fast64_t a_;
    in>>a_;
    modint y(a_);
    x=y;
    return in;
}
ostream& operator<<(ostream &out,const modint& x){
    out<<x.a;
    return out;
}
modint pwr(int_fast64_t a,int_fast64_t b){
    modint _;
    int_fast64_t n=1,A=a;
    while(b){
        if(b&1) (n*=A)%=modint::MOD;
        (A*=A)%=modint::MOD;
        b>>=1;
    }
    _.a=n;
    return _;
}


int main() {
    int n, m;
    cin >> n >> m;

    if(abs(n-m) >= 2) {
        cout << "0" << endl;
    } else if(n == m) {
        modint tmp = 1;
        REPN(i, 1, n+1) {
            tmp *= i;
        }
        REPN(i, 1, m+1) {
            tmp *= i;
        }
        tmp *= 2;
        cout << tmp << endl;
    } else {
        modint tmp = 1;
        REPN(i, 1, n+1) {
            tmp *= i;
        }
        REPN(i, 1, m+1) {
            tmp *= i;
        }
        cout << tmp << endl;        
    }

    


}