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 Rust (1.15.1) (judge verdict: Accepted)

use std::vec::Vec;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

const MOD: usize = 1_000_000_000 + 7;
fn main() {
    let (n, m) = {
        let v: Vec<usize> = read_vec();
        (v[0], v[1])
    };

    if n == m {
       let ans = Mod{ val: 2 } * Mod{ val:  m }.fact() * Mod{ val: n }.fact();
       println!("{}", ans.get_val());
    } else if n + 1 == m || n == m+1 {
       let ans = Mod{ val: m }.fact() * Mod{ val: n }.fact();
       println!("{}", ans.get_val());
    } else {
        println!("0");
    }
}

use std::cmp::*;
use std::ops::Add;
use std::ops::Sub;
use std::ops::Mul;
use std::ops::Neg;
use std::ops::Div;


#[derive(Debug, Clone, Copy, Eq, PartialEq)]
struct Mod {
  val: usize,
}


// 和を定義
impl Add for Mod {
  type Output = Mod;
  // O(1)
  fn add(self, other: Mod) -> Mod {
    Mod { val: (self.val + other.val) % (1_000_000_000 + 7) }
  }
}

// 差を定義
impl Sub for Mod {
  type Output = Mod;

  // O(1)
  fn sub(self, other: Mod) -> Mod {
    let mx = max(self.val, other.val);
    let mn = min(self.val, other.val);

    Mod { val: (mx - mn) % (1_000_000_000 + 7) }
  }
}

// 積を定義
impl Mul for Mod {
  type Output = Mod;

  // O(1)
  fn mul(self, other: Mod) -> Mod {
    Mod { val: (self.val * other.val) % (1_000_000_000 + 7) }
  }
}

// 加法単位元
impl Neg for Mod {
  type Output = Mod;

  // O(1)
  fn neg(self) -> Mod {
    Mod { val: (1_000_000_000 + 7) - self.val }
  }
}

// 逆元を定義
// フェルマーの小定理により実装
impl Div for Mod {
  type Output = Mod;

  //O(logN)
  fn div(self, other: Mod) -> Mod {
    self * other.inv()
  }
}

impl Mod {
  // コンストラクター
  fn new(n: usize) -> Mod {
    Mod { val: n %  (1_000_000_000 + 7) }
  }

  // 加法単位元
  fn new_zero() -> Mod {
    Mod { val: 0 }
  }

  // 乗法単位元
  fn new_one() -> Mod {
    Mod { val: 1 }
  }

  // 加法逆元
  // 上記にて定義



  // 乗法逆元
  fn inv(self) -> Mod {
    if self.val == 0 { panic!("zero divide"); }
    self.pow((1_000_000_000 + 7) - 2)
  }

  // 累乗を定義
  // O(logN)
  fn pow(self, n: usize) -> Mod {
    let mut res = Mod::new(1);
    let mut a = self;
    let mut n = n;

    while n > 0 {
      if n % 2 == 0 {
        a = a * a;
        n /= 2;
      } else {
        res = res * a;
        a = a * a;
        n /= 2;
      }
    }
    res
  }

  // 最大公約数
  // O(logN)
  fn gcd(self, other: Mod) -> usize {
      let mut x = self.val;
      let mut y = other.val;
      while y != 0 {
          let t = y;
          y = x % y;
          x = t;
      }
      x
  }

  // 階乗
  // O(N)
  fn fact(self) -> Mod {
    let mut res = Mod::new_one();
    for i in 1..self.val+1 {
      res = res * Mod::new(i)
    }

    res
  }

  // 順列
  fn perm(self, other: Mod) -> Mod {
    let mut res = Mod::new_one();
    for x in (self.val - other.val + 1)..self.val+1 {
      res = res * Mod::new(x);
    }

    res
  }

  // 組み合わせ O(N)
  fn conb(self, other: Mod) -> Mod {
    self.fact() / (other.fact() * (self - other).fact())
  }

  fn get_val(self) -> usize { self.val }
}