B. The Eternal Immortality

B. The Eternal Immortality

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Even if the world is full of counterfeits, I still regard it as wonderful.

Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.

The phoenix has a rather long lifespan, and reincarnates itself once every a! years. Here a! denotes the factorial of integer a, that is, a! = 1 × 2 × ... × a. Specifically, 0! = 1.

Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of b! years, that is, . Note that when b ≥ a this value is always integer.

As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.

Input

The first and only line of input contains two space-separated integers a and b (0 ≤ a ≤ b ≤ 10¹⁸).

Output

Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.

Examples

Input

2 4

Output

2

Input

0 10

Output

0

Input

107 109

Output

2

Note

In the first example, the last digit of is 2;

In the second example, the last digit of is 0;

In the third example, the last digit of is 2.

题意看样例应该很明白了，现在要做的就是，怎么简化，其实只要比较各位就够了，如果两个数相差的数中

有结尾为零的，就直接是零了，否则就个位数累乘。

#include <iostream>
#define ll long long int
using namespace std;

int main(){
  ll n,m;
  cin>>n>>m;
  int x=n%10;
  int y=m%10;
  if(m-n>=10){
    cout<<0<<endl;
  }else if(y<x){
    cout<<0<<endl;
  }else{
    int cnt=1;
    for(int i=x+1;i<=y;i++){
      cnt*=i;
      cnt=cnt%10;
    }
    cout<<cnt<<endl;
  }
  return 0;
}