hdu2197

f [ n ] = 2 ^ n - f [ k ]; ( k | n )

#include<bits/stdc++.h>
using namespace std;
#define maxn 100000005
typedef long long ll;
int f[maxn];

ll mod_pow(ll x,ll n,int mod)
{
    ll res=1;
    while(n)
    {
        if(n&1)
          res=res*x%mod;
        x=x*x%mod;
        n>>=1;
    }
    return res;
}

int work(ll n)
{
  if(f[n]!=0) return f[n];
  int ans=mod_pow(2,n,2008)-2;
  for(int i=2;i*i<=n;i++)
  {
      if(n%i==0)
      {
        f[i]=work(i)%2008;
        ans=(ans%2008-f[i]+2008)%2008;
        if(i*i!=n)
        {
            f[n/i]=work(n/i)%2008;
            ans=(ans%2008-f[n/i]+2008)%2008;
        }
      }
  }
  return f[n]=ans;
}

int main()
{
    f[1]=2,f[2]=2,f[3]=6,f[4]=12;
    ll n;
    while(~scanf("%lld",&n))
    {
        printf("%d\n",work(n));
    }
    return 0;
}