Calculation 2


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2246    Accepted Submission(s): 958


Problem Description


Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.


 



Input


For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.


 



Output


For each test case, you should print the sum module 1000000007 in a line.


 



Sample Input


3 4 0


 



Sample Output


0 2 


/*
HDOJ 3501 不互质数的和
欧拉函数phi(n)求1-n中与n互质的质因子的个数，
如果gcd(n,i)==1,那么就有gcd(n,n-i)==1;
所以与n互质的数应该是成对出现的，每一对的和就是n,
所以互质总和就是n*phi(n)/2;

于是求小于n并且与n不互质的所有数的和，总和减去上面互质的和  
*/
#include<iostream>
#include<stdio.h>
using namespace std;

int phi(int x)
{
  int i,ans=x;
  for(i=2;i*i<=x;i++)
  {
    if(x%i==0)
    {
      ans-=ans/i;
      while(x%i==0)//i肯定是素数 
        x/=i;
    }
  }
  if(x>1)
    ans=ans/x*(x-1);
  return ans;
}

int main()
{
  __int64 ans,n;
  while(scanf("%I64d",&n),n)
  {
    ans=n*(n+1)/2-n;//总和 
    ans-=phi(n)*n/2;//减去互质的总和公式 
    ans%=1000000007;//再取模 
    printf("%I64d\n",ans);
  } 
  return 0;
}