Codeforces Round #439 (Div. 2) 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 integera, 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 ofb! years, that is, . Note that whenb ≥ 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 is2;

In the second example, the last digit of is0;

In the third example, the last digit of is2.

题意：

         两个数阶乘取商 求商的最后一位

有坑点！！！！（需要longlong类型 会超 int   说多了都是泪~~）

思路  模拟过程就好  不难发现如果两个数字差距超过5  那么第一个数字的阶乘一定是第二个数字阶乘的十倍+  所以商最后一位一定是0

特判下这个就好其他地方 如果 超过五  也会增加十倍 具体可以 模拟1~9的阶乘不难发现 所以 如果两个数字  %=10 之后仍然在5的两侧  那么一定 差十倍 输出0

对于剩下的情况  就可以for来写了~~

#include <iostream>#include <string>#include <cstdio>using namespace std;const int N = 50005;int main(){long long  a, b;scanf("%I64d%I64d",&a,&b);if ((b - a) >= 5 ) {printf("0\n");}else {a %= 10; b %= 10;if (a < 5 && b > 5 || a > 5 && b < 5) {printf("0\n");}else {long long ans = 1;for (int i = a + 1; i <= b; i ++) {ans *= i;}printf("%I64d\n",ans%10);}}}