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

题意：看那些式子也知道了，就是求1-n中有多少对互质的数，即phi[1]+phi[2]+......+phi[n]

题解：预处理出phi，然后求一个前缀和，就完事了~~

预处理主要用到了欧拉函数的性质2和性质3

欧拉函数的三个性质

性质1： 当q为质数时phi(q)=q-1

性质2： 如果i/p是p的倍数时phi(i)=phi(i/p)*p      这里p是最小质因子

性质3： 如果i/p不是p的倍数时phi(i)=phi(i/p)*(p-1)    这里p是最小质因子

性质2，性质3的用法，具体看代码：

上代码：

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
const int MAX = 1e6+520;
ll a[MAX],p[MAX],phi[MAX],sum[MAX];
bool vis[MAX];
int cnt;
void init(){
	vis[1]=vis[0]=1;
	for (int i = 2; i < MAX;i++){
		if(!vis[i]){
			p[cnt++]=i;
			a[i]=i;//a数组存的是最小质因子
		}
		for (int j = 0; j < cnt&&i*p[j] < MAX;j++){
			vis[i*p[j]]=1;
			a[i*p[j]]=p[j];
			if(i%p[j]==0) break;
		}
	}
}
void getphi(){
	phi[1]=1;
	for (int i = 2; i < MAX;i++){
		if(a[i/a[i]]==a[i]) phi[i]=phi[i/a[i]]*a[i];//性质2
		else phi[i]=phi[i/a[i]]*(a[i]-1);//性质3
	}
}
int main(){
	init();
	getphi();
	for (int i = 2; i < MAX;i++) sum[i]=sum[i-1]+phi[i];
	int n;
	while(scanf("%d",&n)&&n) printf("%lld\n",sum[n]);
	return 0;
}