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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <functional>
#include <math.h>
#include <stdlib.h>
#include <map>
#include <deque>
#include <sys/timeb.h>
#include <fstream>

using namespace std;

#define repr(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repr(i,0,n)
#define reprrev(i,a,b) for(int i=(int)(b)-1;i>=(int)(a);i--)
#define reprev(i,n) reprrev(i,0,n)
#define repi(itr,ds) for(auto itr=ds.begin();itr!=ds.end();itr++)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define mp make_pair
#define mt make_tuple
#define INF 1050000000
#define INFL 1100000000000000000LL
#define EPS (1e-10)
#define MOD 1000000007
#define PI 3.1415926536
#define RMAX 4294967295

typedef long long ll;
typedef pair<int, int> Pi;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<Pi> vPi;
typedef vector<vector<int> > vvi;
typedef vector<vector<bool> > vvb;
typedef vector<vector<ll> > vvll;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<double> > vvd;
typedef vector<vector<Pi> > vvPi;
typedef priority_queue<int, vector<int>, greater<int> > pqli;
typedef priority_queue<ll, vector<ll>, greater<ll> > pqlll;
typedef priority_queue<Pi, vector<Pi>, greater<Pi> > pqlP;
struct Edge {
	int from, to, cost;
	bool operator<(Edge e) {
		return cost < e.cost;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
template <class T>
using vec = vector<T>;
template<class T>
using pql = priority_queue<T, vector<T>, greater<T>>;
string debug_show(Pi a) {
	return "(" + to_string(a.first) + "," + to_string(a.second) + ")";
}

template<class T>
struct augEdge {
	T from, to;
	int cost;
	bool operator<(augEdge e) { return cost < e.cost; }
	bool operator>(augEdge e) { return cost > e.cost; }
};
template<class T>
using augGraph = vector<augEdge<T>>;


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n, m;
	cin >> n >> m;
	if (abs(n - m) >= 3) {
		cout << 0 << endl;
		return 0;
	}
	ll ans1 = 1, ans2 = 1;
	repr(i, 1, n+1) {
		ans1 = (ans1 * i) % MOD;
	}
	repr(i, 1, m+1) {
		ans2 = (ans2 * i) % MOD;
	}
	ll ans;
	if (abs(n - m) % 2 == 0) {
		ans = (ans1 * ans2 * 2) % MOD;
	}
	else {
		ans = (ans1 * ans2) % MOD;
	}
	
	cout << ans << endl;
	
	return 0;
}