Farey Sequence

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<cstdio>
#include<cstring>
long long int f[1000010];
int main()
{
    int n;
    memset(f,0,sizeof(f));
    for(int i=2; i<1000010; i++)
    {
        if(!f[i])
        {
            for(int j=i; j<1000010; j+=i)
            {
                if(!f[j])
                    f[j]=j;
                f[j]=f[j]/i*(i-1);
            }
        }
    }
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        long long int ans=0;
        for(int i=2; i<=n; i++)
        {
            ans+=f[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}