This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.

An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.

As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.

For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.

Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it's the product of three H-primes.

Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.

For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.

#include <iostream>

#include <string.h>

using namespace std;

const int MAXN=;

bool isHprime[MAXN];

int Hprime[MAXN],top;

int h[MAXN];

void sieve()

{

    memset(isHprime,true,sizeof(isHprime));

    for(int i=;i<MAXN;i+=)

    {

        if(isHprime[i])

        {

            Hprime[top++]=i;

            for(int j=i+i;j<MAXN;j+=i)

            {

                if((j-)%==)

                {

                    isHprime[j]=false;

                }

            }

        }

    }

    for(int i=;i<top;i++)

    {

        if(Hprime[i]>)    break;

        for(int j=i;j<top;j++)

        {

            int mul=Hprime[i]*Hprime[j];

            if(mul>=MAXN)

            {

                break;

            }

            h[mul]=;

        }

    }

    for(int i=;i<MAXN;i++)

    {

        h[i]+=h[i-];

    }

}

int main()

{

    sieve();

    int n;

    while(cin>>n&&n!=)

    {

        cout<<n<<" "<<h[n]<<endl;

    }

    return ;

}