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 <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <set>
#include <utility>
#include <cstdlib>
#include <queue>
#include <stack>
#include <iomanip>
#include <cstdio>
#include <map>

using namespace std;

typedef long long Int;
typedef pair<Int,Int> P;

#define rep(i,x) for(Int i = 0; i < (Int)(x); i++)
#define rrep(i,x) for(Int i = ((Int)(x) - 1); i >= 0; i--)
#define _upgrade ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define all(x) (x).begin(), (x).end()
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define pb push_back


template<class T1, class T2> void chmin(T1 &a, T2 b){if(a>b)a=b;}
template<class T1, class T2> void chmax(T1 &a, T2 b){if(a<b)a=b;}

//ll gcd(ll a, ll b){return b?gcd(b,a%b):a;}
//ll lcm(ll x, ll y) {return x / gcd(x, y) * y;}

Int dx[4] = {1,0,-1,0};
Int dy[4] = {0,1,0,-1};
const Int INF = 1001002003004005006ll;
const Int mod = 1e9 + 7;
const double PI = 3.14159265358979323846;


// ****************************************CODE***************************************//






Int fact[100000000];


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

        Int tmp = max(n,m);
        fact[0] = 1;
        for(Int i = 1; i <= tmp; i++){
                fact[i] = fact[i-1] * i;
                fact[i] %= mod;
        }

        if(n == m){
                Int tmp = fact[n];
                tmp *= tmp;
                tmp %= mod;
                tmp *= 2;
                tmp %= mod;
                cout << tmp << endl;
                return 0;
        }

        if(abs(m-n) > 1){
                cout << 0 << endl;
                return 0;
        }

        Int tmp1 = fact[n];
        Int tmp2 = fact[m];

        Int ans = tmp1 * tmp2 % mod;

        cout << ans << endl;
        return 0;
}