A. Vasya and Petya's Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya and Petya are playing a simple game. Vasya thought of number x between 1 and n, and Petya tries to guess the number.

Petya can ask questions like: "Is the unknown number divisible by number y?".

The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.

Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers yi, he should ask the questions about.

Input

A single line contains number n (1 ≤ n ≤ 103).

Output

Print the length of the sequence of questions k (0 ≤ k ≤ n), followed by k numbers — the questions yi (1 ≤ yi ≤ n).

If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.

Sample test(s)

input

4

output

32 4 3 

input

6

output

42 4 3 5 

Note

The sequence from the answer to the first sample test is actually correct.

If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.

If the unknown number is divisible by 4, it is 4.

If the unknown number is divisible by 3, then the unknown number is 3.

Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.

解题说明：此题是给出１－ｎ内的一个数ｋ，然后输出一个询问序列ｙi, 在询问完后可以判断出这个数为什么。每次询问为ｋ是否为yi的倍数。做法是从因子角度看ｋ，我们只要确定ｋ中质因子的种类以及对应的个数即可。看ｋ中有多少个２，３，５。

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cmath>using namespace std;int tmp[10000]={0};int main () {int n;scanf("%d",&n);int ans=0;for(int i=2;i<=n;i++) {if(tmp[i]==0) {for(int j=i+i;j<=n;j+=i) {tmp[j]++;}}}for(int i=2;i<=n;i++) {if(tmp[i]<2){ans++;}}printf("%d\n",ans);for(int i=2;i<=n;i++){if(tmp[i]<2){printf("%d ",i);}}return 0;}