#include <bits/stdc++.h>
#define rep(i, p, n) for (ll i = p; i < (ll)(n); i++)
#define rep2(i, p, n) for (ll i = p; i >= (ll)(n); i--)
using namespace std;
using ll = long long;
using ld = long double;
const double pi = 3.141592653589793;
const long long inf = 2 * 1e9;
const long long linf = 4 * 1e18;
const ll mod1 = 1000000007;
const ll mod2 = 998244353;
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}

// atcoder
#include <atcoder/all>
using namespace atcoder;
using mint1 = modint1000000007;
using mint2 = modint998244353;

vector<pair<ll, ll>> base = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

//素数列挙
//output: (void)N以下の整数のうち素数であるものをsosuuに返す
void SR(ll N, vector<ll> &sosuu) { 
    vector<bool> testSR(N+1, true);
    for(int i=2; i<=N; i++) {
        if (testSR.at(i)) {
            sosuu.push_back(i);
            for(int j=i*2; j<=N; j+=i) {
                testSR.at(j)=false;
            }
        }
    }
}

int main()
{

    //////////////////
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    //////////////////

    ll N;
    cin >> N;
    vector<ll> sosuu(0);
    SR(N, sosuu);
    ll M = sosuu.size();
    vector<ll> co(M);
    ll now = 0;
    for (ll c : sosuu)
    {
        //N!中の素因数cの個数
        ll temp = N;
        while(temp!=0) {
            co.at(now) += temp / c;
            temp /= c;
        }
        now++;
    }
    vector<ll> li(M, 1);
    vector<vector<ll>> kake(2000000);
    rep(i, 0, M) {
        rep(j, 0, co.at(i)) {
            kake.at(co.at(i)/(j+1)).push_back(i);
            //cout << co.at(i)/(j+1) << " " << i << endl;
        }
    }
    mint2 ans = 0, now2=1, B=1;
    rep2(i, 1999999, 0) {
        for(ll c:kake.at(i)) {
            now2 /= li.at(c);
            li.at(c)++;
            now2 *= li.at(c);
        }
        ans += (now2-B) * i;
        B=now2;
    }
    cout << ans.val();
}