Problem:
Score : 300 points

Problem Statement

You are given an integer N.
Find the number of the positive divisors of N!, modulo 10^9+7.

Constraints

1≤N≤10^3

Input

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

N

Output

Print the number of the positive divisors of N!, modulo 10^9+7.

Sample Input 1

3

Sample Output 1

4

There are four divisors of 3! =6: 1, 2, 3 and 6. Thus, the output should be 4.

Sample Input 2

6

Sample Output 2

30

Sample Input 3

1000

Sample Output 3

972926972

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

#![allow(unused_imports, unused_variables, dead_code)]
use std::io::*;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

trait InputValue {
    fn parse(s: &str) -> Self;
}

fn read<T: InputValue>() -> T {
    let mut buf = String::new();
    let _ = stdin().read_line(&mut buf);
    T::parse(&buf.trim())
}

fn readnc<T: InputValue>() -> Vec<T> {
    let mut vec = vec![];
    let line: String = read();
    for token in line.split_whitespace() {
        vec.push(T::parse(token));
    }
    vec
}

fn readn<T: InputValue>(n: usize) -> Vec<T> {
    let mut vec = vec![];
    for _ in 0..n {
        vec.push(read());
    }
    vec
}

macro_rules! parse_single_value {
        ($($t:ty),*) => {
            $(
                impl InputValue for $t {
                    fn parse(s: &str) -> $t { s.parse().unwrap() }
                }
            )*
    	}
    }
parse_single_value!(i32, i64, f32, f64, usize, String);

macro_rules! parse_tuple {
    	($($t:ident),*) => {
    		impl<$($t),*> InputValue for ($($t),*) where $($t: InputValue),* {
    			fn parse(s: &str) -> ($($t),*) {
    				let mut tokens = s.split_whitespace();
    				let t = ($($t::parse(tokens.next().unwrap())),*);
    				t
    			}
    		}
    	}
    }
parse_tuple!(A, B);
parse_tuple!(A, B, C);

// ===

const MOD: i64 = 1000000007;

fn main() {
    let n: usize = read();
    let mut ans: i64 = 1;
    let mut isp: Vec<bool> = vec![true; 1010];
    for i in 2..isp.len() {
        if !isp[i] {
            continue
        }

        let mut cantake = 0;
        let mut nn = n;
        while nn >= 1 {
            cantake += nn / i;
            nn /= i;
        }
        ans *= (cantake + 1) as i64;
        ans %= MOD;


        let mut ii = i * i;
        while ii < isp.len() {
            isp[ii] = false;
            ii += i;
        }
    }
    println!("{}", ans);
}