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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <vector>

#define FOR(i,k,n) for (int (i)=(k); (i)<(n); ++(i))
#define rep(i,n) FOR(i,0,n)
#define pb push_back
#define all(v) begin(v), end(v)
#define debug(x) cerr<< #x <<": "<<x<<endl
#define debug2(x,y) cerr<< #x <<": "<< x <<", "<< #y <<": "<< y <<endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef vector<ll> vll;
typedef vector<vector<ll> > vvll;
template<class T> using vv=vector<vector< T > >;

struct Prime {
  int max_prime;
  int prime_number;
  vector<int> primes;
  map<int, int> prime_id;
  Prime() {};
  Prime(int m) {
    initialize(m);
  };
  void initialize(int m) {
    sieve(m);
    prime_number = (int)primes.size();
  }

  void sieve(int m) {
    max_prime = 2;
    deque<bool> is_prime(m+1, true);
    is_prime[0] = is_prime[1] = false;
    int i;
    for (i = 2; i <= m; ++i) {
      if (!is_prime[i]) {
        continue;
      }
      if (i * i > m) {
        break;
      }
      for (int j = i+1; j <= m; ++j) {
        if (j % i == 0) {
          is_prime[j] = false;
        }
      }
      prime_id[i] = (int)primes.size();
      primes.push_back(i);
      max_prime = i;
    }
    for (int j = i; j <= m; ++j) {
      if (!is_prime[j]) {
        continue;
      }
      prime_id[j] = (int)primes.size();
      primes.push_back(j);
      max_prime = j;
    }
  }

  map<int, int> prime_factor(ll x) {
    map<int, int> ret;
    if ((ll)max_prime * (ll)max_prime < x) {
      return ret;
    }
    for (int i = 0; i < prime_number; ++i) {
      while (x % primes[i] == 0) {
        x /= primes[i];
        ret[primes[i]] += 1;
      }
    }
    if (x != 1) {
      ret[x] += 1;
    }
    return ret;
  }
};

int mod = 1e9+7;

ll po(ll k, ll x) {
  if (x == 0) {
    return 1;
  }
  if (x == 1) {
    return k % mod;
  }
  ll y = po(k, x/2);
  y = y * y % mod;
  if (x % 2 == 1) {
    y = y * k % mod;
  }
  return y;
}

int main() {
  int n;
  scanf("%d", &n);
  Prime pr(32);

  map<int, int> mps;
  FOR (i, 2, n+1) {
    auto appends = pr.prime_factor(i);
    for (auto append : appends) {
      mps[append.first] += append.second;
    }
  }
  ll ans = 1;
  for (auto m : mps) {
    ans = ans * (m.second + 1) % mod;
  }
  printf("%lld\n", ans);

  return 0;
}