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 9.2.1) (judge verdict: Wrong Answer)

    /**   Created by: Marcos Portales
          Codeforces User: marcosportales
          Country: Cuba
          Copyright                    */
#include<bits/stdc++.h>
using namespace std;
/// Pragmas:
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline") //Optimization flags
#pragma GCC option("arch=native","tune=native","no-zero-upper") //Enable AVX
#pragma GCC target("avx2") //Enable AVX
/// Macros:
#define int long long
#define f first
#define s second
#define db(x) cerr << #x << ": " << (x) << '\n';
#define pb push_back
#define lb lower_bound
#define up upper_bound
#define all(x) x.begin() , x.end()
#define rall(x) x.rbegin() , x.rend()
#define enl '\n'
#define vi vector<int>
#define sz(a) int((a).size())
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,x,y) for(int i=(x);i<=(y);i++)
typedef pair<int,int> ii;
typedef long double ld;
typedef unsigned long long ull;
/// Constraints:
const int maxn = 400010;
const int mod = 1000000007;
const ld eps = 1e-9;
const int inf = ((1ll<<31ll)-1ll);
const int INF = 2000000000000000000ll;
const ld pi = acos(-1);
/// Prime Numbers:
// 2, 173, 179, 181, 197, 311, 331, 737, 1009, 2011, 2027, 3079, 4001, 10037, 10079, 20011, 20089;
// 100003, 100019, 100043, 200003, 200017, 1000003, 1000033, 1000037, 1000081;
// 2000003, 2000029, 2000039, 1000000007, 1000000021, 2000000099;
/// Functions:
#define lg2(x) 31 - __builtin_clz(x)
#define lgx(x,b) ( log(x) / log(b) )
/// Red-Black Tree Test.insert(x)late ---------------------------------
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree < long long ,  null_type ,  less<long long> ,  rb_tree_tag ,  tree_order_statistics_node_update > ordered_set;
/// Quick Pow------------------------------------------------
int qpow(int b,int e){
    if( !e ) return 1;
    if( e & 1 ) return qpow(b,e-1) * b % mod;
    int pwur = qpow(b,e>>1);
    return (pwur * pwur) % mod;
}
int modinv(int x){
    return qpow(x,mod-2);
}
/// My Code -------------------------------------------------

int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cout.setf(ios::fixed); cout.precision(0);
    srand(time(NULL));
    //freopen("a.in","r",stdin);
    //freopen("a.in","w",stdout);
    int n,m;
    cin>>n>>m;
    int ans=1;
    REP(i,2,n){
        ans=(ans*i)%mod;
    }
    REP(i,2,m){
        ans=(ans*i)%mod;
    }
    if(n==m){
        cout<<ans*2%mod;
    }else if(n<m){
        cout<<0;
    }else cout<<ans<<endl;

    return 0;
}