Calculation 2


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3141    Accepted Submission(s): 1300



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


//题意：


求小于且不与n互质的数之和。


用欧拉函数先求出与n互质的数之和，再用总和减去就行了。


Hint： 欧拉函数求得的与n互质的值为n*ephi(n)/2; 


#include<stdio.h>
#include<string.h>
#include<math.h>
#define M 1000000007
long long ephi(long long n)
{
	int m=(int)sqrt(n+0.5);
	long long ans=n;
	for(int i=2;i<=m;i++)
		if(n%i==0)
		{
			ans=ans/i*(i-1);
			while(n%i==0)
				n/=i;
		}
		if(n>1)
			ans=ans/n*(n-1);
		return ans;
}
int main()
{
	long long n;
	while(scanf("%lld",&n),n)
	{
		long long ans=((n*(n-1)/2)-((n*ephi(n)/2)))%M;
		printf("%lld\n",ans);
	}
	return 0;
}