CF822A.I'm bored with life

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?

假期结束了。在黑客Leha的帮助下，Noora成功进入了她梦寐以求的大学——帕夫洛波利斯（Pavlopolis）大学。众所周知，大学会为学生在校学习期间提供宿舍。因此，Noora不得不离开维奇科波利斯（Vičkopolis），搬往帕夫洛波利斯。于是，Leha彻底独自一人留在了宁静的小镇维奇科波利斯，甚至因百无聊赖而差点陷入抑郁！

为了稍微放松一下，Leha给自己出了一个任务：他任选两个整数 $A$ 和 $B$，然后计算整数 “$A$ 的阶乘” 与 “$B$ 的阶乘” 的最大公约数。形式上，这位黑客希望求出 $\mathrm{GCD}(A!,\, B!)$。众所周知，整数 $x$ 的阶乘是所有不大于 $x$ 的正整数之积，即 $x! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (x - 1) \cdot x$。例如，$4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24$。回忆一下，$\mathrm{GCD}(x,\, y)$ 是能同时整除（无余数）$x$ 和 $y$ 的最大正整数 $q$。

Leha 已经学会了非常高效地解决这一问题。你也能做到，而且不会比他差，对吧？

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

第一行且唯一一行包含两个整数 $A$ 和 $B$（$1 \leq A, B \leq 10^9$，$\min(A, B) \leq 12$）。

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

输出一个整数，表示整数 $A!$ 和 $B!$ 的最大公约数。

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.

考虑以下示例：

$4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24$，$3! = 1 \cdot 2 \cdot 3 = 6$。整数 $24$ 和 $6$ 的最大公约数恰好为 $6$。