#include<bits/stdc++.h>
#define inf 0x3f3f3f3f3f3f3f3fll
#define debug(x) cerr<<#x<<"="<<x<<endl
using namespace std;
using ll=long long;
using ld=long double;
using pli=pair<ll,int>;
using pi=pair<int,int>;
template<typename A>
using vc=vector<A>;
inline int read()
{
    int s=0,w=1;char ch;
    while((ch=getchar())>'9'||ch<'0') if(ch=='-') w=-1;
    while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
    return s*w;
}
inline ll lread()
{
    ll s=0,w=1;char ch;
    while((ch=getchar())>'9'||ch<'0') if(ch=='-') w=-1;
    while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
    return s*w;
}
const int mod=998244353;
int pri[400005];
bool is[400005];
ll g[2][400005];
ll f1[400005];
ll f2[400005];
ll sq,n;
inline ll qow(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
inline ll& gp(ll w)
{
    if(!w) return g[0][0];
    if(w<=sq) return g[0][w];
    return g[1][n/w];
}
inline ll get(ll m)
{
    ll v=gp(m)-gp(m/2)+1;
    if(m==1) return 1;
    if(v==m) return qow(m%mod,m-2);
    //共 v 个孤立点和 1 个连通块
    // printf("m=%lld v=%lld\n",m,v);
    return (m-v)%mod*qow(m%mod,v-1)%mod;
}
int main()
{
    n=lread();
    for(;(sq+1)*(sq+1)<=n;sq++);
    // printf("sq=%lld\n",sq);

    for(int i=2;i<=sq;i++)
    {
        if(!is[i]) pri[++pri[0]]=i;
        for(int j=1;j<=pri[0]&&i*pri[j]<=sq;j++)
        {
            is[i*pri[j]]=1;
            if(i%pri[j]==0) break;
        }
    }
    
    vc<ll>num;
    for(ll i=1;i<=sq;i++) num.push_back(i);
    for(ll i=sq;i;i--) num.push_back(n/i);
    num.erase(unique(num.begin(),num.end()),num.end());
    for(ll i:num) gp(i)=i-1;
    
    unsigned sw=0;
    // for(ll i:num) printf("%lld ",i);;putchar('\n');
    for(int i=1;i<=pri[0];i++)
    {
        while(sw<num.size()&&(ll)pri[i]*pri[i]>num[sw]) sw++;
        for(unsigned j=num.size()-1;j>=sw;j--) gp(num[j])-=gp(num[j]/pri[i])-(i-1);
    }

    // debug(G(11,2));
    // debug(G(3,1)-G(2,1));
    // for(ll i=1;i<=n;i++) if(i==1||n/i!=n/(i+1)) printf("%lld : %lld\n",n/i,fp(n/i));

    ll ans=1;
    for(ll l=1,r;l<=n;l=r+1)
    {
        r=n/(n/l);
        // printf("l=%lld r=%lld\n",l,r);
        ans=ans*qow(get(n/l),r-l+1)%mod;
        // printf("ans=%lld\n",ans);
    }
    printf("%lld\n",ans);
    return 0;
}