#include <bits/extc++.h>
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

#define depref(x, y) (x[y] - x[y-1])

// To use most b i ts rather than jus t the lowest ones :
struct chash { // large odd number for C
    const uint64_t C = ll(4e18 * acos(0)) | 71;

    ll operator()(ll x) const { return __builtin_bswap64(x*C); }
};

__gnu_pbds::gp_hash_table<ll,ll,chash> s_f({},{},{},{},{1<<25});

const ll mod = 998244353;
const ll CAP = 20210926;
const ll DPC = 100000;

vector<ll> dp(DPC + 1, 0);

void prep(){
    dp[0] = 0;
    dp[1] = 1;

    for(int i = 2; i <= DPC; i++){
        for(int j = 1; j * j <= i; j++){
            if(i % j)
                continue;

            dp[i] += dp[i / j];

            if(j * j != i)
                dp[i] += dp[j];

            dp[i] %= mod;
        }
    }

    for(int i = 1; i <= DPC; i++)
        dp[i] = (dp[i] + dp[i - 1]) % mod;
}

ll solve(ll n){
    if(n <= DPC)
        return dp[n];
    
    if(s_f.find(n) != s_f.end())
        return s_f[n];
    
    ll ans = 0;
    for(ll i = 1, ub; i <= n; i = ub + 1){
        ub = n / (n / i);
        
        ans += solve(n/i) * (min(CAP, ub) - i + 1) % mod;
        ans %= mod;
    }

    return s_f[n] = ans;
}

int main(){
    ll n;
    cin >> n;

    prep();

    cout << solve(n) << "\n";
}