Three Digits

Per is obsessed with factorials. He calculates them, estimates them, reads about them, draws them, and dreams about them. The value $12! = 479001600$ is tattooed on his back.

He noticed long ago that a factorial ends with several zeros, and he wrote a program that counts those trailing zeros. For example $12!$ ends with 600, so it has 2 trailing zeros. Now he wants to look one step further, at the 3 digits that sit right before the trailing zeros. Removing the trailing zeros from $12!$ leaves 4790016, whose last 3 digits are 016.

You are given an integer $n$. Remove every trailing zero from $n!$ and find the last 3 digits of what remains. If fewer than 3 digits remain, find all of them.

The first line contains one integer $n$ ($1 \le n \le 10^7$).

Print, on one line, the last 3 digits of $n!$ after every trailing zero is removed. Print a leading 0 as it is. If the remaining value is 4032, print 032. If the remaining value has fewer than 3 digits, print that value as it is.