HDU 3501 Calculation 2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MOD=1000000007 ;
/*
小于n与n互质的数的和为Eular（n）×n/2
*/
long long Eular(long long n){
    long long res=n;
    for(int i=2;i*i<=n;i++){
        if(n%i==0){
            res=res/i*(i-1);
            while(n%i==0)
                n/=i;
        }
    }
    if(n>1)
        res=res/n*(n-1);
    return res;
}

int main(){
    long long n;
    while(cin>>n&&n){
        long long ans=n*(n+1)/2-n;
        ans=ans-Eular(n)*n/2;
        cout<<ans%MOD<<endl;
    }
    return 0;
}