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.*;
import java.util.*;

public class Main {
  private static long solve(int n, int m) {
    if (Math.abs(n - m) > 1) {
      return 0;
    }
    ModCalculator mc = new ModCalculator(1_000_000_007L);
    long answer = mc.mul(mc.getF(n), mc.getF(m));
    if (n == m) {
      return mc.mul(answer, 2);
    }
    return answer;
  }
  
  private static void execute(ContestReader reader, PrintWriter out) {
    int n = reader.nextInt();
    int m = reader.nextInt();
    out.println(solve(n, m));
  }
  
  public static void main(String[] args) {
    ContestReader reader = new ContestReader(System.in);
    PrintWriter out = new PrintWriter(System.out);
    execute(reader, out);
    out.flush();
  }
}

class ContestReader {
  private BufferedReader reader;
  private StringTokenizer tokenizer;
  
  ContestReader(InputStream in) {
    reader = new BufferedReader(new InputStreamReader(in));
  }
  
  public String next() {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      try {
        tokenizer = new java.util.StringTokenizer(reader.readLine());
      } catch (Exception e) {
        throw new RuntimeException(e);
      }
    }
    return tokenizer.nextToken();
  }
  
  public int nextInt() {
    return Integer.parseInt(next());
  }
  
  public long nextLong() {
    return Long.parseLong(next());
  }
  
  public double nextDouble() {
    return Double.parseDouble(next());
  }
  
  public String[] next(int n) {
    String[] array = new String[n];
    for (int i = 0; i < n; i++) {
      array[i] = next();
    }
    return array;
  }
  
  public int[] nextInt(int n) {
    int[] array = new int[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextInt();
    }
    return array;
  }
  
  public long[] nextLong(int n) {
    long[] array = new long[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextLong();
    }
    return array;
  }
  
  public double[] nextDouble(int n) {
    double[] array = new double[n];
    for (int i = 0; i < n; i++) {
      array[i] = nextDouble();
    }
    return array;
  }
  
  public char[] nextCharArray() {
    return next().toCharArray();
  }
  
  public int[][] nextInt(int n, int m) {
    int[][] matrix = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextInt();
      }
    }
    return matrix;
  }
  
  public long[][] nextLong(int n, int m) {
    long[][] matrix = new long[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextLong();
      }
    }
    return matrix;
  }
  
  public double[][] nextDouble(int n, int m) {
    double[][] matrix = new double[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = nextDouble();
      }
    }
    return matrix;
  }
  
  public char[][] nextCharArray(int n) {
    char[][] matrix = new char[n][];
    for (int i = 0; i < n; i++) {
      matrix[i] = next().toCharArray();
    }
    return matrix;
  }
}

class ModCalculator {
  private final long mod;
  private final ModCombinationCache modCombinationCache;
  
  ModCalculator(long mod) {
    this.mod = mod;
    this.modCombinationCache = new ModCombinationCache();
  }
  
  public long add(long a, long b) {
    return (a + b) % mod;
  }
  
  public long sub(long a, long b) {
    return (a - b + mod) % mod;
  }
  
  public long mul(long a, long b) {
    return (a * b) % mod;
  }
  
  public long pow(long a, long b) {
    if (b == 0) {
      return 1;
    }
    long v = pow(mul(a, a), b / 2);
    if (b % 2 == 1) {
      return mul(v, a);
    } else {
      return v;
    }
  }
  
  public long inverse(long a) {
    return pow(a, mod - 2);
  }
  
  public long div(long a, long b) {
    return mul(a, inverse(b));
  }
  
  public long getF(int n) {
    return modCombinationCache.getF(n);
  }
  
  public long getP(int n, int r) {
    return modCombinationCache.getP(n, r);
  }
  
  public long getC(int n, int k) {
    return modCombinationCache.getC(n, k);
  }
  
  class ModCombinationCache {
    private final List<Long> factorialCache;
    private final List<Long> factorialInverseCache;
    
    public ModCombinationCache() {
      factorialCache = new ArrayList<>();
      factorialCache.add(1L);
      factorialInverseCache = new ArrayList<>();
      factorialInverseCache.add(1L);
    }
    
    private void resize(int n) {
      for (int i = factorialCache.size() - 1; i < n; i++) {
        long v = mul(factorialCache.get(i), i + 1);
        factorialCache.add(v);
        factorialInverseCache.add(inverse(v));
      }
    }
    
    long getF(int n) {
      resize(n);
      return factorialCache.get(n);
    }
    
    long getP(int n, int r) {
      resize(n);
      return mul(factorialCache.get(n), factorialInverseCache.get(n - r));
    }
    
    long getC(int n, int k) {
      resize(n);
      return mul(factorialCache.get(n), mul(factorialInverseCache.get(k), factorialInverseCache.get(n-k)));
    }
  }
}