Codeforces Round #439 (Div. 2) 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.

题目大意：给你a和b，求b的阶乘除于a的阶乘的结果的最后一位

刚开始写的时候我竟然傻傻的暴力了一发.......T了以后仔细想想其实没有那么难，这应该算是一道思维题，我们一般会想到，既然b大于a，那么a那一部分肯定是不用算的，所以我们算b-a就行了，如果b-a大于等于10的话，那么阶乘里面肯定有一个10，也就是说，得出来的答案的最后一位肯定是0，连算都不用算，其他的话就暴力就行了

#include<iostream>using namespace std;typedef long long ll;int main(){ll a,b;cin>>a>>b;int ans=1;if(!a&&!b){cout<<1<<endl;return 0;}else{if(b-a>=10) cout<<0<<endl;else{for(ll i=a+1;i<=b;i++){ans*=(i%10);if(ans/10) ans%=10;}cout<<ans<<endl;}}}