A. I'm bored with life

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies.
Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!

Leha came up with a task for himself to relax a little. He chooses two integers A and B and
then calculates the greatest common divisor of integers "A factorial" and "B factorial".
Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is
a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x.
For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is
the largest positive integer q that divides (without a remainder) both x and y.

Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?

Input

The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).

Output

Print a single integer denoting the greatest common divisor of integers A! and B!.

Example

input

4 3

output

6

Note

Consider the sample.

4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6.
The greatest common divisor of integers 24 and 6 is
exactly 6.

解题说明：此题是求两个数的最大公约数，不过和一般的不同，这里的两个数求的是连乘结果，很显然，大的数连乘肯定包含小的数的连乘，最小的那个数的连乘结果就为这两个数的最大公约数了。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<iostream>
using namespace std;

int main()
{
int a,b,x,i,p=1;
scanf("%d %d",&a,&b);
x=a<b?a:b;
for(i=1;i<=x;i++)
{
p=p*i;
}
printf("%d\n",p);
return 0;
}