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 PyPy3 (7.3.0) (judge verdict: Accepted)

import math
from math import gcd,pi,sqrt
INF = float("inf")

import sys
sys.setrecursionlimit(10**6)
import itertools
from collections import Counter,deque
def i_input(): return int(input())
def i_map(): return map(int, input().split())
def i_list(): return list(i_map())
def i_row(N): return [i_input() for _ in range(N)]
def i_row_list(N): return [i_list() for _ in range(N)]
def s_input(): return input()
def s_map(): return input().split()
def s_list(): return list(s_map())
def s_row(N): return [s_input for _ in range(N)]
def s_row_str(N): return [s_list() for _ in range(N)]
def s_row_list(N): return [list(s_input()) for _ in range(N)]


def main():
    N,M = i_map()
    s = abs(N-M)
    z = 10**9 + 7

    if s >= 2:
        print(0)
        exit()

    if s == 0:
        a = math.factorial(N)
        print(a**2*2%z)
    else:
        a = math.factorial(N)
        b = math.factorial(M)
        print(a*b%z)

if __name__=="__main__":
    main()