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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws IOException {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskX solver = new TaskX();
		solver.solve(1, in, out);
		out.close();
	}

	static int INF = 1 << 30;
	static int MOD = 1000000007;
	static int[] mh4 = { 0, -1, 1, 0 };
	static int[] mw4 = { -1, 0, 0, 1 };

	static class TaskX {
		public void solve(int testNumber, InputReader in, PrintWriter out) {

			int n = in.nextInt(), m = in.nextInt();
			if (Math.abs(n - m) > 1) {
				out.println(0);
				return;
			}

			if (n == m) {
				out.println(fact[n] % MOD * fact[m] % MOD * 2 % MOD);
			} else {
				out.println(fact[n] % MOD * fact[m] % MOD);
			}

		}
	}

	/**
	 * 二項係数
	 * 前提 n < modP
	 * nCr = n!/(r!*(n-r)!)である。この時分子分母にMODが来る場合は以下のように使用する
	 * */
	public static long comb(int n,int r) {
		return fact[n] * factInv[r] % MOD * factInv[n-r] % MOD;
	}

	/**
	 * 階乗数の逆元
	 *
	 * */
	public static int MAXN = 200000;

	static long[] fact = factorialArray(MAXN, MOD);
	static long[] factInv = factorialInverseArray(MAXN, MOD, inverseArray(MAXN, MOD));

	// 階乗の mod P テーブル
	public static long[] factorialArray(int maxN,long mod) {
		long[] fact = new long[maxN+1];
		fact[0] = 1 % mod;
		for(int i=1;i<=maxN;i++) {
			fact[i] = fact[i-1] * i % mod;
		}
		return fact;
	}
	// 数 i に対する mod P での逆元テーブル
	public static long[] inverseArray(int maxN,long modP) {
		long[] inv = new long[maxN+1];
		inv[1] = 1;
		for(int i=2;i<=maxN;i++) {
			inv[i] = modP - (modP / i) * inv[(int) (modP%i)] % modP;
		}
		return inv;
	}

	// 階乗の逆元テーブル
	public static long[] factorialInverseArray(int maxN,long modP,long[] inverseArray) {
		long[] factInv = new long[maxN+1];
		factInv[0] = 1;
		for(int i=1;i<=maxN;i++) {
			factInv[i] = factInv[i-1] * inverseArray[i] % modP;
		}
		return factInv;
	}


	static class InputReader {
		BufferedReader in;
		StringTokenizer tok;

		public String nextString() {
			while (!tok.hasMoreTokens()) {
				try {
					tok = new StringTokenizer(in.readLine(), " ");
				} catch (IOException e) {
					throw new InputMismatchException();
				}
			}
			return tok.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(nextString());
		}

		public long nextLong() {
			return Long.parseLong(nextString());
		}

		public double nextDouble() {
			return Double.parseDouble(nextString());
		}

		public int[] nextIntArray(int n) {
			int[] res = new int[n];
			for (int i = 0; i < n; i++) {
				res[i] = nextInt();
			}
			return res;
		}

		public long[] nextLongArray(int n) {
			long[] res = new long[n];
			for (int i = 0; i < n; i++) {
				res[i] = nextLong();
			}
			return res;
		}

		public InputReader(InputStream inputStream) {
			in = new BufferedReader(new InputStreamReader(inputStream));
			tok = new StringTokenizer("");
		}

	}

}