Sherlock and his girlfriend

B. Sherlock and his girlfriend

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.

He bought n pieces of jewelry. The i-th piece has price equal to i + 1, that is, the prices of the jewelry are 2, 3, 4, ... n + 1.

Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.

Help Sherlock complete this trivial task.

Input

The only line contains single integer n (1 ≤ n ≤ 100000) — the number of jewelry pieces.

Output

The first line of output should contain a single integer k, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.

The next line should consist of n space-separated integers (between 1 and k) that specify the color of each piece in the order of increasing price.

If there are multiple ways to color the pieces using k colors, you can output any of them.

Examples

input

3

output

21 1 2 

input

4

output

22 1 1 2

Note

In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.

In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.

这道题我一点也不想说啥，英语不好是硬伤啊，我没看到prime这个单词，所以其难度可想而知，就是陈立杰也很难做啊，看完人家的代码才知道原来就是是素数输出1，非素数输出2，就这么简单。

#include<stdio.h>int n;int book[100010];int main(){int i,j;scanf("%d",&n);for(i=2;i<=n+1;i++){for(j=2*i;j<=n+1;j+=i)book[j]=1; //说明j不是素数 }if(n<=2) //也就是2 或者 2,3这两种情况 {printf("1\n");for(i=1;i<=n;i++)  printf("1 ");printf("\n");return 0;}printf("2\n");for(i=1;i<=n;i++){if(book[i+1]==1) //素数输出1 {printf("1 ");}else       //非素数输出2 printf("2 ");}printf("\n");return 0;}