Farey Sequence

Description

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are 
F2 = {1/2} 
F3 = {1/3, 1/2, 2/3} 
F4 = {1/4, 1/3, 1/2, 2/3, 3/4} 
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5} 

You task is to calculate the number of terms in the Farey sequence Fn.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10 ⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn. 

Sample Input

2
3
4
5
0

Sample Output

1
3
5
9

#include<stdio.h>
#include<string.h>
__int64 e[1000005],ans[1000005];
void fun()  ///筛法求
{
	int i,j;
	for(i=2;i<1000005;i++)
		if(!e[i])
			for(j=i;j<1000005;j+=i)
			{
				if(!e[j])
					e[j]=j;
				e[j]=e[j]-e[j]/i;
			}
	for(i=1;i<1000005;i++)
		ans[i]=ans[i-1]+e[i];
}
int main()
{
	int n;
	fun();
	while(~scanf("%d",&n)&&n)
		printf("%I64d
",ans[n]);
}