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

核心思想：

欧拉函数前n项和。 详见代码。

代码如下：

#include<cstdio> #include<iostream> using namespace std; typedef long long ll; const int N=1e6+40; int ou[N]; ll ans[N]; void oula()//将欧拉函数打表 { for(int i=2;i<N;i++) { if(ou[i])continue; for(int j=i;j<N;j+=i) { if(!ou[j]) ou[j]=j; ou[j]=ou[j]/i*(i-1); } } return; } int main() { oula(); //求欧拉函数的前n项和 for(int i=2;i<N;i++) ans[i]=ans[i-1]+ou[i]; int n; while(1) { scanf("%d",&n); if(!n) break; printf("%lld\n",ans[n]); } return 0; }