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>
#define for0(i, n) for(int i = 0; i < n; i++)
#define for1(i, n) for(int i = 1; i <= n; i++)
#define pb push_back
#define mp make_pair
#define all(v) v.begin(), v.end()
#define V vector<int>
#define VP vector<pair<int, int> >
#define clr(A, x) memset(A, x, sizeof(A))
#define cpy(A, B) memcpy(A, B, sizeof(B))
#define g(s) getline(cin, s) ///ai grija la fin/cin ///
#define FASTIO ios_base::sync_with_stdio(0)
const long long INFLL = 2 * 1e18 + 100;
const int INFINT = 2 * 1e9 + 100;
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
//ifstream fin("a2.in");
//ofstream fout("a2.out");
void die()
{
    cout << "-1";
    exit(0);
}
const int NMAX = 1e5 + 5;
const int MOD = 1e9 + 7; /// careful here (7 or 9, 66.. etc)
const double PI = atan(1) * 4;
const double EPS = 1e-12;

ll fact[NMAX];


int main()
{
    FASTIO;
    int n, m;
    cin >> n >> m;
    fact[0] = 1;
    for1(i, NMAX - 1) fact[i] = i * fact[i - 1] % MOD;
    if(n == m)
    {
        cout << 2 * fact[n] * fact[n] % MOD;
    }
    else if(abs(n - m) == 1) cout << fact[n] * fact[m] % MOD;
    else cout << 0;


    return 0;
}