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: Wrong Answer)

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

//#include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision;
// using cint = cpp_int;
//#define input(a) scanf("%lld", &(a))
//#define output(a) printf("%lld\n", (a))

// Define
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const ll MOD = 1e9 + 7;
const ll INF = LONG_MAX;
const ull MAX = ULONG_MAX;
#define endl '\n'
#define space ' '
#define def inline auto
#define func inline constexpr ll
#define run __attribute__((constructor)) def _
#define all(v) begin(v), end(v)

// Debug
#define debug(...)                                                             \
    {                                                                          \
        cerr << __LINE__ << ": " << #__VA_ARGS__ << " = ";                     \
        for (auto &&X : {__VA_ARGS__}) cerr << "[" << X << "] ";               \
        cerr << endl;                                                          \
    }

// Loop
#define inc(i, a, n) for (ll i = (a), _##i = (n); i <= _##i; ++i)
#define dec(i, a, n) for (ll i = (a), _##i = (n); i >= _##i; --i)
#define each(i, a) for (auto &&i : a)
#define rep(i, n) inc(i, 0, n - 1)

// Stream
#define fout(n) cout << fixed << setprecision(n)
#define fasten cin.tie(0), ios::sync_with_stdio(0)

// Speed
run() { fasten, fout(10); }
#pragma GCC optimize("O3")
#pragma GCC optimization_level 3
#pragma GCC target("avx")

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

def input() {
    ll A;
    cin >> A;
    return A;
}

ll N, M, S, L;
vector<vector<bool>> A(1000);

signed main() {
    ll n, m, a = 1, b = 1;
    cin >> n >> m;
    if(abs(n - m) > 1) {
        cout << 0 << endl;
        return 0;
    }
    rep(i, n) if(i) { a *= i; a %= MOD; }
    rep(i, m) if(i) { b *= i; b %= MOD; }
    cout << a * b << endl;
}

// for compilation: g++ -Ofast -march=native -o _ _.cpp -std=c++17