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: Accepted)

#include<bits/stdc++.h>

using namespace std;

#define LL long long
#define DD double
#define Pb push_back
#define Bp pop_back
#define Pf push_front
#define Fp pop_front
#define Ub upper_bound
#define Lb lower_bound
#define In insert
#define All(x) x.begin(), x.end()
#define mem(a, b) memset(a, b, sizeof(a))
#define fast ios_base::sync_with_stdio(0);cin.tie(0)
#define X first
#define Y second
#define mx1 10
#define mx2 110
#define mx3 1010
#define mx4 10010
#define mx5 100010
#define mx6 1000010

typedef vector<LL> Vl;
typedef vector<DD> Vd;
typedef vector<bool> Vb;
typedef vector<Vl> VVl;
typedef pair<LL, LL> Pll;
typedef pair<DD, DD> Pdd;
typedef vector<Pll> Vpl;
typedef vector<Pdd> Vpd;
typedef queue<LL> Ql;
typedef stack<LL> Sl;
typedef deque<LL> Dl;

LL _set(LL N, LL pos) {return N = N | (1 << pos);}
LL _reset(LL N, LL pos) {return N = N & ~(1 << pos);}
bool _check(LL N, LL pos) {return (bool)(N & (1 << pos));}
bool _upper(char a) {return a >= 'A' && a <= 'Z';}
bool _lower(char a) {return a >= 'a' && a <= 'z';}
bool _digit(char a) {return a >= '0' && a <= '9';}

LL dx[] = {1, -1, 0, 0, -1, -1, 1, 1};
LL dy[] = {0, 0, 1, -1, -1, 1, -1, 1};

///**************************DONE****************************///

LL mod = 1e9 + 7;

LL fact(LL n)
{
    if(n < 2)
        return 1;

    LL ans = 1;
    for(int i = 2; i <= n; i++)
        ans = (i * ans) % mod;
    return ans;
}

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

    if(abs(n - m) > 1)
        return cout << 0 << '\n', 0;

    n = fact(n), m = fact(m);
    LL ans = (n * m) % mod;
    if(n == m)
        ans = (ans * 2) % mod;

    cout << ans << '\n';
    return 0;
}