#include<cstdio>
#include<iostream>
#include<cmath>
#define int long long
#define RI register int
#define CI const int&
using namespace std;
const int mod=998244353;
namespace Min25_Sieve
{
	const int N=1e6+5;
	int n,lim,pri[N],cnt,vis[N],id1[N],id2[N],g[N],w[N],idx;
	inline int ID(CI x)
	{
		return x<=lim?id1[x]:id2[n/x];
	}
	inline void init(CI nn)
	{
		n=nn; lim=(int)sqrt(n); 
		for (RI i=2;i<=lim;++i)
		{
			if (!vis[i]) pri[++cnt]=i;
			for (RI j=1;j<=cnt&&i*pri[j]<=lim;++j)
			if (vis[i*pri[j]]=1,i%pri[j]==0) break;
		}
		for (RI l=1,r;l<=n;l=r+1)
		{
			r=n/(n/l); w[++idx]=n/l;
			if (n/l<=lim) id1[n/l]=idx;
			else id2[n/(n/l)]=idx;
			g[idx]=w[idx]-1;
		}
		for (RI j=1;j<=cnt;++j)
		for (RI i=1;i<=idx;++i)
		{
			if (pri[j]*pri[j]>w[i]) break;
			g[i]-=g[ID(w[i]/pri[j])]-(j-1);
		}
	}
	inline int work(CI n)
	{
		return n<=1?0:g[ID(n)];
	}
};
using namespace Min25_Sieve;
inline int quick_pow(int x,int p=mod-2,int mul=1)
{
	for (;p;p>>=1,x=1LL*x*x%mod) if (p&1) mul=1LL*mul*x%mod; return mul;
}
inline int solve(CI m)
{
	if (m<=2) return 1;
	if (m==3) return 3;
	int tot=work(m)-work(m/2);
	return quick_pow(m%mod,tot)*(m%mod-(tot+1)%mod+mod)%mod;
}
signed main()
{
	int n; scanf("%lld",&n);
	init(n); int ans=1;
	for (RI l=1,r;l<=n;l=r+1)
	r=n/(n/l),(ans*=quick_pow(solve(n/l),r-l+1))%=mod;
	return printf("%lld",ans),0;
}