Problem: You're given an integer n. For every integer i from 2 to n, assign a positive integer a_i such that the following conditions hold:

  * For any pair of integers (i,j), if i and j are coprime, a_i ≠ a_j. 
  * The maximal value of all a_i should be minimized (that is, as small as possible). 



A pair of integers is called [coprime](https://en.wikipedia.org/wiki/Coprime_integers) if their [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) is 1.

Input

The only line contains the integer n (2 ≤ n ≤ 10^5).

Output

Print n-1 integers, a_2, a_3, …, a_n (1 ≤ a_i ≤ n). 

If there are multiple solutions, print any of them.

Examples

Input


4


Output


1 2 1 

Input


3


Output


2 1

Note

In the first example, notice that 3 and 4 are coprime, so a_3 ≠ a_4. Also, notice that a=[1,2,3] satisfies the first condition, but it's not a correct answer because its maximal value is 3.

Solution (in PYTHON):

Solution:
# -*- coding: utf-8 -*-
import sys,fractions,bisect
from math import factorial
def fi():
    return int(sys.stdin.readline())

def fi2():
    return map(int, sys.stdin.readline().split())

def fi3():
    return sys.stdin.readline()

def fo(*args):
    for s in args:
        sys.stdout.write(str(s)+' ')
    sys.stdout.write('\n')
INF=10**9+7
sys.setrecursionlimit(INF)

#main
n=fi()
ans=[-1 for i in range(n+1)]
p=0
for i in range(2,n+1):
    #print "i=",i
    #print "root i=",int(i**(0.5))
    c=0
    for j in range(2,int(i**(0.5))+1):
        if i%j==0:
            ans[i]=ans[j]
            c=1
            break
    if c==0:
        p+=1
        ans[i]=p

res=[]
for i in range(2,len(ans)):
    res.append(str(ans[i]))

fo(" ".join(res))