#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N=1e6+100;
const int M=1e6+5;
const ll mod=998244353;
ll p[M];
int cnt;
ll getmi(ll x,ll y){
	ll ret=1;
    x%=mod;
	while(y){
		if(y&1) ret=ret*x%mod;
		x=x*x%mod;
		y=y>>1;
}
return ret;
}
const ll inv2=getmi(2,mod-2);
const ll inv4=getmi(4,mod-2);
const ll inv6=getmi(6,mod-2);
bitset<N> v;
void seive()
{
    for(int i=2;i<N;++i)
    {
        if(!v[i])
            p[++cnt]=i;
        for(int j=1;j<=cnt && i*p[j]<N;++j)
        {
            v[i*p[j]]=true;
            if(i%p[j]==0)
                break;
        }
    }
}
struct min25
{
    ll n,sum0[M],sum1[M],sum2[M],sum3[M];
    ll w[N<<1],g0[N<<1],g1[N<<1],g2[N<<1],g3[N<<1],G[N<<1];
    int id1[N],id2[N],tot,sq;
    ll calc(ll j,ll x)//f(p^x)的值 
	{
		return p[j]^x;
	}

    int getid(ll x)
    {
        if(x<=sq)
            return id1[x];
        return id2[n/x];
    }
    void init(ll nn)
    {
        for(int i=1;i<=cnt;++i)
        {
            sum0[i]=(sum0[i-1]+1);
            // sum1[i]=(sum1[i-1]+p[i])%mod;
            // sum2[i]=(sum2[i-1]+p[i]*p[i]%mod)%mod;
            // sum3[i]=(sum3[i-1]+p[i]*p[i]%mod*p[i]%mod)%mod;
        }
        tot=0;
        n=nn;
        sq=sqrt(n);
        for(ll r,mm,m,l=1;l<=n;l=r+1)
        {
            r=n/(n/l);

            w[tot]=m=mm=n/l;
            
            g0[tot]=(mm-1);
            // g1[tot]=mm*(mm+1)%mod*inv2%mod-1;
            // g2[tot]=mm*(mm+1)%mod*(2*mm%mod+1)%mod*inv6%mod-1;
            // g3[tot]=mm*mm%mod*(mm+1)%mod*(mm+1)%mod*inv4%mod-1;
            if(m<=sq)
                id1[m]=tot;
            else
                id2[n/m]=tot;
            ++tot;
        }
        for(int i=1;i<=cnt;++i)
        {
            for(int j=0;j<tot && p[i]*p[i]<=w[j];++j)
            {
                ll t=getid(w[j]/p[i]);
                g0[j]=(g0[j]-(g0[t]-sum0[i-1]));
                // g1[j]=(g1[j]-p[i]*(g1[t]-sum1[i-1])%mod)%mod;
                // g2[j]=(g2[j]-p[i]*p[i]%mod*(g2[t]-sum2[i-1])%mod)%mod;
                // g3[j]=(g3[j]-p[i]*p[i]%mod*p[i]%mod*(g3[t]-sum3[i-1])%mod)%mod;
            }
        }
        // rep(i,0,tot-1) G[i]=((g1[i]-g0[i])%mod+mod)%mod;//所有的K 均为calc(1) g[i]即质数的0次方和
		// repd(j,cnt,1)
		// 	rep(i,0,tot-1)
		// 	{
		// 		if((ll)p[j]*p[j]>w[i]) break;
		// 		for(ll k=p[j],c=1; k*p[j]<=w[i]; ++c,k*=p[j],(G[i]+=(p[j]^c))%=mod)
		// 			(G[i]+=(p[j]^c)%mod*(G[getid(w[i]/k)]-(g1[getid(p[j])]-g0[getid(p[j])])%mod+mod)%mod)%=mod;
		// 	}

    }
    ll get(ll x)
	{
		return G[getid(x)]+1;
	}
    ll cal0(ll r)
    {
        return (g0[getid(r)]);
    }
    ll cal1(ll r)
    {
        return (g1[getid(r)])%mod;
    }
    ll cal2(ll r)
    {
        return (g2[getid(r)])%mod;
    }
    ll cal3(ll r)
    {
        return (g3[getid(r)])%mod;
    }
}m2;

ll ksm(ll x,ll k){
    ll res=1;
    while(k){
        if(k&1)res=res*x%mod;
        x=x*x%mod;
        k/=2;
    }
    return res;
}
int main()
{
    seive();
    ll n;cin>>n;
    m2.init(n);
    ll ans=1;
    auto solve = [&](ll n){
        ll res=1;
        if(n==1)return res;
        ll p=m2.cal0(n)-m2.cal0(n/2)+1;
        // assert(p>=2);
        if(n-p>0)res=ksm(n%mod,p+1-2);
        else res=ksm(n%mod,p-2);
        if(n-p>0)res=res*((n-p)%mod)%mod;
        // cout<<"n="<<n<<" res="<<res<<" p="<<p<<endl;
        return res;
    };
    for(ll l=1,r;l<=n;l=r+1){
        r=min(n/(n/l),n);
        ll x=n/l;
        ans=ans*ksm(solve(x),r-l+1)%mod;
    }
    cout<<ans;
}