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>
#include <unordered_map>

const double PI = acos(-1);

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
using P = pair<ll, ll>;
using V = vector<ll>;
#define rep(i, n) for(ll i=0;i<(ll)(n);i++)
#define rep1(i, n) for(ll i=1;i<=(ll)(n);i++)
#define rep2(i, m, n) for(ll i=m;i<(ll)(n);i++)
#define rrep(i, n, m) for(ll i=n;i>=(ll)(m);i--)
#define ALL(obj) (obj).begin(), (obj).end()
#define RALL(obj) (obj).rbegin(), (obj).rend()
#define MOD 1000000007
#define INF 1000000000

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 1000000007;
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x% mod + mod) % mod) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) {
		(x *= a.x) %= mod;
		return *this;
	}
	mint operator+(const mint a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint a) const {
		mint res(*this);
		return res *= a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const {
		return pow(mod - 2);
	}
	mint& operator/=(const mint a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const {
		mint res(*this);
		return res /= a;
	}
};

// nCr
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
// usage:
//   combination comb(n+5);
//   ans = comb(n,r);
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		assert(n < mod);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
};

// 階乗
mint factorial(ll k) {
	mint sum = 1;
	for (mint i = 1; i.x <= k; i = i + 1)
	{
		sum *= i;
	}
	return sum;
}

void solve_abc_c() {
	ll n, m;
	ll a, b;
	//ll ans = 0;
	//double ans = 0;
	
	cin >> n >> m;

	mint ans;
	mint tmp = 2;

	if (abs(n - m) > 1) {
		ans = 0;
	}
	else {
		if (n == m) {
			ans = factorial(n) * factorial(n) * tmp;
		}
		else {
			ans = factorial(max(m, n)) * factorial(min(m, n));
		}
	}

	cout << ans.x << "\n";
}

int main(void) {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed;
	cout << setprecision(15);

	solve_abc_c();

	return 0;
}