Codeforces Round #400 (Div. 1 + Div. 2, combined)B.Sherlock and his girlfriend【预处理素数+思维】

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.

题目大意：

第i个数表示i+1的价值，其中价值为x和价值为y的如果x是y的因子，那么两个位子的颜色不能相同。

问用最少的颜色来涂所有的格子，怎样分配。

思路：

#include<stdio.h>#include<string.h>using namespace std;int vis[10000800];int ans[10000800];int main(){    int n;    while(~scanf("%d",&n))    {        if(n<3)        {            printf("1\n");            for(int i=1;i<=n;i++)printf("1 ");            printf("\n");        }        else        {            memset(ans,0,sizeof(ans));            memset(vis,0,sizeof(vis));            for(int i=2;i<=100005;i++)            {                if(!vis[i])                {                    ans[i]=1;                    for(int j=2;j*i<=100005;j++)                    {                        vis[i*j]=1;                    }                }            }            printf("2\n");            for(int i=2;i<=n+1;i++)            {                if(ans[i]==1)printf("1 ");                else printf("2 ");            }            printf("\n");        }    }}

预处理出1e5之内的所有素数，将素数处理过程中的基数设定为颜色1，其他的数字设定为颜色2即可。

那么就用两种颜色就能完成这个目的。

Ac代码：