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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;;
#define MEM(a,b) memset((a),(b),sizeof(a))
const LL INF = 1e9 + 7;
const int N = 2e2 + 10;
int getsum(int n, int x)
{
	int ret = 0;
	while (n)
	{
		n /= x;
		ret += n;
	}
	return ret;
}
bool check(int n)
{
	for (int i = 2; i * i <= n; i++)
	{
		if (n % i == 0) return false;
	}
	return true;
}
int main()
{
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	int n;
	cin >> n;
	LL ans = 1;
	for (int i = 2; i <= n; i++)
	{
		if (!check(i)) continue;
		ans = ans * (getsum(n, i) + 1) % INF;
	}
	cout << ans << endl;
	return 0;
}