Problem:
Score : 200 points

Problem Statement

Snuke loves working out. He is now exercising N times.

Before he starts exercising, his power is 1. After he exercises for the i-th time, his power gets multiplied by i.

Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.

Constraints

1 ≤ N ≤ 10^{5}

Input

The input is given from Standard Input in the following format:

N

Output

Print the answer modulo 10^{9}+7.

Sample Input 1

3

Sample Output 1

6

After Snuke exercises for the first time, his power gets multiplied by 1 and becomes 1.

After Snuke exercises for the second time, his power gets multiplied by 2 and becomes 2.

After Snuke exercises for the third time, his power gets multiplied by 3 and becomes 6.

Sample Input 2

10

Sample Output 2

3628800

Sample Input 3

100000

Sample Output 3

457992974

Print the answer modulo 10^{9}+7.

Solution in Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::io;
#[allow(unused_imports)]
use std::cmp;
#[allow(unused_imports)]
use std::collections::LinkedList;

fn read_list() -> LinkedList<String> {
    let mut s = String::new();
    io::stdin().read_line(&mut s).expect("");
    s.split(' ').map(|x| x.to_string()).collect::<LinkedList<String>>()
}

fn pop_parse<T: std::str::FromStr>(list: &mut LinkedList<String>) -> T {
    list.pop_front().expect("").trim().parse::<T>().ok().unwrap()
}


fn main() {
    let modulo = 1e9 as i64 + 7;
    let mut n = read_list();
    let n: i64 = pop_parse(&mut n);
    let mut pow:i64 = 1;
    for i in 1..n+1 {
        pow = (pow * i) % modulo;
    }
    println!("{}", pow);
}