H

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

看几个例子可以发现，后一个总是比前一个多这次的数与之前的数互质的个数，用欧拉就可以，因为是1e6，所以要用筛法的，不能用直接的；

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
typedef long long ll;
typedef long double ld;
const ll mod=1e9+7;
using namespace std;
const double pi=acos(-1.0);
ll euler[1000005];
void chuli()
{
	 euler[1]=1;
     for(int i=2;i<1000001;i++)
       euler[i]=i;
     for(int i=2;i<1000001;i++)
        if(euler[i]==i)
           for(int j=i;j<1000001;j+=i)
              euler[j]=euler[j]/i*(i-1);//先进行除法是为了防止中间数据的溢出
	for(int i=3;i<1000001;i++)
	 euler[i]=euler[i]+euler[i-1];
}
    
int main()
{
	//freopen("output1.txt", "r", stdin);
	chuli();
	int n;
	while(1)
	{
		sf("%d",&n);
		if(n==0) return 0;
		pf("%lld
",euler[n]);
	} 
}