#include<bits/stdc++.h>

using namespace std;
using ll = long long;
const int mo = 998244353;
std::vector<int> minp, primes;
void sieve(int n) {
    minp.assign(n + 1, 0);
    primes.clear();
    for (int i = 2; i <= n; i++) {
        if (minp[i] == 0) {
            minp[i] = i;
            primes.push_back(i);
        }
        for (auto p : primes) {
            if (i * p > n) break;
            minp[i * p] = p;
            if (p == minp[i]) {
                break;
            }
        }
    }
}
int qp(int a,int b) {
    int ans = 1,base = a;
    while(b != 0) {
        if(b & 1) ans = 1ll * ans * base % mo;
        base = 1ll * base * base % mo;
        b >>= 1;
    }
    return ans;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n;
    cin >> n;
    ll s = sqrt(n);
    sieve(s);
    vector<ll> v;
    for(ll l = 1,r;l <= n;l = r + 1) {
        v.push_back(n / l);
        r = n / (n / l);
    }

    vector<ll> f1(v.size()),f2(v.size());
    auto getid = [&](ll x) -> ll {
        if(x <= s) return v.size() - x;
        else return n / x - 1;
    };
    ll inv2 = qp(2,mo - 2),inv6 = qp(6,mo - 2);
    for(int i = 0;i < v.size();i ++) {
        ll x = v[i];
        x %= mo;
        f1[i] = ((x + 1) * x % mo * inv2 % mo + mo - 1) % mo;// \sum_2^n i^k = \sum_1^n i^k - 1
        f2[i] = (x * (x + 1) % mo * (2 * x % mo + 1) % mo * inv6 % mo + mo - 1) % mo;
    }
    ll ps1 = 0,ps2 = 0;
    for(auto p : primes) {
        for(int i = 0;i < v.size();i ++) {
            if(1ll * p * p > v[i]) break;
            f1[i] -= 1ll * p * (f1[getid(v[i] / p)] - ps1 + mo) % mo;
            f1[i] += mo;f1[i] %= mo;
            f2[i] -= 1ll * p * p % mo * (f2[getid(v[i] / p)] - ps2 + mo) % mo;
            f2[i] += mo;f2[i] %= mo;
        }
        ps1 += p;
        ps1 %= mo;
        ps2 += 1ll * p * p % mo;
        ps2 %= mo;
    }
    // cerr << f1[getid(n)] << " " << f2[getid(n)] << endl;
    ll F1 = (f1[getid(n)] - f1[getid(n / 2)] + mo) % mo;
    ll F2 = (f2[getid(n)] - f2[getid(n / 2)] + mo) % mo;
    n %= mo;
    ll sum1 = (2 + n) % mo * (n - 1) % mo * inv2 % mo;
    ll sum2 = (n % mo * (n + 1) % mo * (2 * n % mo + 1) % mo * qp(6,mo - 2) % mo + mo - 1) % mo;




    // cerr << F1 << " " << F2 << endl;
    ll ans = (sum1 * F1 % mo - (F1 * F1 % mo - F2 + mo) % mo * inv2 % mo + mo - F2 + mo) % mo;
    // cerr << ans << endl;
    ll Ans = (sum1 * sum1 % mo - sum2 + mo) % mo * inv2 % mo;
    // cerr << Ans << endl;
    ans = (Ans - ans + mo) % mo;
    cout << ans << endl; 
}