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)

use std::env;
use std::io::*;
use std::str::*;
use std::mem;
use std::collections::HashMap;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let s = stdin.bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok().unwrap()
}
fn read_char() -> char {
    let c = read::<String>();
    let c = c.chars().collect::<Vec<char>>();
    c[0]
}
fn read_String_Vec() -> Vec<char> {
    let S = read::<String>();
    S.chars().collect::<Vec<char>>()
}

fn vecchar_to_String(Vs: Vec<char>) -> String {
    let mut S = String::new();
    for i in 0..Vs.len() {
        S.push(Vs[i]);
    }
    S
}

/*
A : 65
Z : 90
a : 97
z : 122
*/

/*
let N = read::<i32>();
let S = read::<String>();
let S = read_String_Vec();
let C = read_char();
*/

/*
 .to_string() : * -> String
 &*S          : String -> &str
*/

fn main() {
    // main の引数
    /*
        引数なしのときargs.len() = 1;
    */
    let args: Vec<String> = env::args().collect();

    let N = read::<i64>();
    let mut mpl = 1;
    for i in 0..N {
        mpl = (mpl * (i+1)) % 1000000007;
    }
    println!("{}", mpl%1000000007);

}