#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long LL;
const int maxn = 1e6 + 5, mod = 998244353;
bool isPrime[maxn];
int cnt;
int id1[maxn], id2[maxn];
LL primes[maxn], w[maxn], g1[maxn], sum1[maxn], g2[maxn], sum2[maxn];
LL n, sq;

void init(){
    for(int i = 2; i <= sq; i++){
        if (!isPrime[i]) primes[++cnt] = i;
        for(int j = 1; i * primes[j] <= sq; j++){
            isPrime[i * primes[j]] = 1;
            if (i % primes[j] == 0) break;
        }
    }
    for(int i = 1; i <= cnt; i++){
        sum1[i] = (sum1[i - 1] + primes[i]) % mod;
        sum2[i] = (sum2[i - 1] + primes[i] * primes[i]) % mod;
    }
}

inline LL f1(LL x){
    x %= mod;
    return x * (x + 1) / 2 % mod;
}

inline LL f2(LL x){
    x %= mod;
    return x * (x + 1) % mod * (2 * x + 1) % mod * 166374059 % mod;
}

inline LL f3(LL x){
    return f1(x) * f1(x) % mod;
}

pair<LL, LL> prime_sum(LL t){

    auto get = [&](LL x){
        return x <= sq ? id1[x] : id2[t / x];
    };
    memset(g1, 0, sizeof g1);
    int m = 0;
    for(LL l = 1, r; l <= t; l = r + 1){
        // 离散化
        r = t / (t / l), w[++m] = t / l;
        // 修改4: 初始化g为多项式的前缀和,需要排除f(1),
        // 注意多项式前的系数必须为1,否则就不是完全积性函数.比如f(x) = k,必须转化为f(x) = 1.
        g1[m] = f1(w[m]) - 1, g2[m] = f2(w[m]) - 1;
        if (w[m] <= sq) id1[w[m]] = m;
        else id2[t / w[m]] = m;
    }
    for(int i = 1; i <= cnt; i++){
        for(int j = 1; j <= m && primes[i] * primes[i] <= w[j]; j++){
            // 修改5: g[j - 1] -> g[j]
            g1[j] = (g1[j] - primes[i] * (g1[get(w[j] / primes[i])] - sum1[i - 1]) % mod + mod) % mod;
            g2[j] = (g2[j] - primes[i] * primes[i] % mod * (g2[get(w[j] / primes[i])] - sum2[i - 1]) % mod + mod) % mod;
        }
    }
    return make_pair(g1[get(t)], g2[get(t)]);
}

int main(){
#ifdef LOCAL
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
#endif

    // 结论,如果两个数字(u,v),存在某个数字是大于n/2的质数,则没有贡献,否则有贡献
    // 所以求出所有大于n/2质数的和与平方和就可以计算了.
    scanf("%lld", &n);
    sq = sqrt(n);
    init();
    auto [s1, s2] = prime_sum(n);
    auto [s3, s4] = prime_sum(n / 2);
    LL sum = s1 - s3;
    if (sum < 0) sum += mod;
    LL sum2 = s2 - s4;
    if (sum2 < 0) sum2 += mod;
    LL t = (sum * (f1(n) - 1)) % mod;
    if (t < 0) t += mod;
    t -= (sum * sum - sum2) % mod * ((mod + 1) / 2) % mod;
    t = (t % mod + mod) % mod;
    t -= sum2;
    if (t < 0) t += mod;
    LL all = (f3(n) - f2(n)) * ((mod + 1) / 2) % mod - (f1(n) - 1);
    all = (all % mod + mod) % mod;
    LL ans = all - t;
    cout << (ans % mod + mod) % mod << '\n';

}