Sherlock and his girlfriend

關於題目

Sherlock and his girlfriend

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.

Example

Input

3

Output

2
1 1 2 

Input

4

Output

2
2 1 1 2

題目大意

有n個珠寶，價值依次爲2,3….n+1,現在要給每個珠寶塗色，要求是：若x是y的因子，則價值爲x所對應的珠寶不能和價值爲y的珠寶塗相同的顏色
其實最多有兩種方案，當n<=2的時候只有一種方案，都塗一樣的顏色

代碼

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const int Maxn = 1e5;
int a[Maxn + 5];
int main()
{
    memset(a, 0, sizeof a);
    int n;
    scanf("%d",&n);
    if(n <= 2)
    {
        printf("1\n");
        for(int i = 0; i < n-1; ++i)
            printf("1 ");
        printf("1");
    }
    else
    {
        for(int i = 2; i <= n+1; ++i)
        {
            for(int j = i+i; j <= n+1; j += i)
                a[j] = 2;
        }
        printf("2\n");
        for(int i = 2; i <= n+1; ++i)
        {
            if(a[i])
                printf("2");
            else
                printf("1");
            if(i <= n)
                printf(" ");
        }
    }
    return 0;
}