#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using pll = pair<ll, ll>;

#define fir first
#define sec second
#define mpr make_pair

#define vv vector
#define eb emplace_back

#define Fr(i, l, r) for (ll i = l; i <= r; ++i)
#define Rf(i, r, l) for (ll i = r; i >= l; --i)

template<typename T> void Min(T& x, T y) {
	if (y < x) x = y;
}

template<typename T> void Max(T& x, T y) {
	if (y > x) x = y;
}

const int N = 1000010;
const long long MD = 998244353, inv2 = (MD + 1) / 2;

ll n, sqrt_n;
ll p[N];
bitset<N> npr;
ll mindiv[N];
ll G[N];
int id1[N], id2[N];

ll sum1(ll n) {
	// n %= MD;
	return ((__int128)n * (n + 1) / 2 - 1) % MD;
}

ll sum2(ll n) {
	// n %= MD;
	return ((__int128)n * (n + 1) * (n * 2 + 1) / 6 - 1) % MD;
}

void sieve() {
	ll lmt = sqrt_n + 100;
	Fr (i, 2, lmt) {
		if (!npr[i]) p[++p[0]] = i, mindiv[i] = i;
		Fr (j, 1, p[0]) {
			if (i * p[j] > lmt) break;
			npr[i * p[j]] = true;
			mindiv[i * p[j]] = p[j];
			if (i % p[j] == 0) break;
		}
	}
	p[0] = 1;
}

void calc() {
	sqrt_n = floor(sqrt(n)), sieve();
	vv<ll> val(1, 0);
	for (ll l = 1, v, r; l <= n; l = r + 1) {
		v = n / l, r = n / v, val.eb(v);
	}
	int cnt = (int)val.size() - 1;
	Fr (i, 1, cnt) {
		G[i] = sum1(val[i]);
		if (val[i] > sqrt_n) id2[n / val[i]] = i;
		else id1[val[i]] = i;
	}
	ll sum = 0;
	Fr (i, 1, sqrt_n) {
		if (p[i] * p[i] > n) break;
		for (ll j = 1; j <= cnt && p[i] * p[i] <= val[j]; ++j) {
			ll tmp = val[j] / p[i];
			ll k = tmp > sqrt_n ? id2[n / tmp] : id1[tmp];
			G[j] = ((G[j] - (G[k] - sum) * p[i]) % MD + MD) % MD;
		}
		sum += p[i];
	}
}

void calc2() {
	vv<ll> val(1, 0);
	for (ll l = 1, v, r; l <= n; l = r + 1) {
		v = n / l, r = n / v, val.eb(v);
	}
	int cnt = (int)val.size() - 1;
	Fr (i, 1, cnt) {
		G[i] = sum2(val[i]);
		if (val[i] > sqrt_n) id2[n / val[i]] = i;
		else id1[val[i]] = i;
	}
	ll sum = 0;
	Fr (i, 1, sqrt_n) {
		if (p[i] * p[i] > n) break;
		for (ll j = 1; j <= cnt && p[i] * p[i] <= val[j]; ++j) {
			ll tmp = val[j] / p[i];
			ll k = tmp > sqrt_n ? id2[n / tmp] : id1[tmp];
			G[j] = ((G[j] - (G[k] - sum) * p[i] % MD * p[i]) % MD + MD) % MD;
		}
		sum = (sum + p[i] * p[i]) % MD;
	}
}

int main() {
	cin >> n;
	calc();
	ll sum = (G[1] - G[2]) % MD;
	ll ans = (sum1(n) - sum) % MD;
	ans = (ans * ans) % MD;
	calc2();
	ll s2 = (G[1] - G[2]) % MD;
	ans = (ans + s2 - sum2(n)) % MD;
	ans = ans * inv2 % MD;
	ans = (ans + MD) % MD;
	cout << ans << '\n';
	return 0;
}