Description

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.

Input

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.

Output

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.

Sample Input

21 85 789 0 Sample Output

21 0 85 5 789 62

大致題意： 一個(gè)H-number是所有的模四余一的數(shù)。 如果一個(gè)H-number是H-primes 當(dāng)且僅當(dāng)它的因數(shù)只有1和它本身（除1外）。 一個(gè)H-number是H-semi-prime當(dāng)且僅當(dāng)它只由兩個(gè)H-primes的乘積表示。 H-number剩下其他的數(shù)均為H-composite。 給你一個(gè)數(shù)h,問1到h有多少個(gè)H-semi-prime數(shù)。

直接打表。。。 有一個(gè)問題，我到現(xiàn)在才發(fā)現(xiàn)if(-1)的判斷也為真。。。非0即為真，寫了兩年C++連這個(gè)都不清楚，真雞兒丟人