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 PYTHON3):

Solution:
# cook your dish here
# cook your dish here
from sys import stdin, stdout
import math,sys
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import bisect as bi
import heapq 
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')): 
    #------------------Sublime--------------------------------------#
    sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
    def I():return (int(input()))
    def In():return(map(int,input().split()))
else:
    #------------------PYPY FAst I/o--------------------------------#
    def I():return (int(stdin.readline()))
    def In():return(map(int,stdin.readline().split()))

def dict(a):
    d={}
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d
def SieveOfEratosthenes(n): 
      
    # Create a boolean array "prime[0..n]" and initialize 
    # all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n + 1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * 2, n + 1, p): 
                prime[i] = False
        p += 1
    prime[0]= False
    prime[1]= False
    # Print all prime numbers 
    ans={}
    for p in range(n + 1): 
        if prime[p]: 
            ans[p]=1
    return ans

def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bi.bisect_right(a, x)
    if i != len(a):
        return i
    else:            
        return -1

def main():
    try:
        d=SieveOfEratosthenes(100000)
        n=I()
        dp=[1e7]*(n)
        j=0
        for x in range(2,n+1):
            if d.get(x,-1)!=-1:
                j+=1
                for i in range(x,n+1,x):
                    dp[i-1]=min(dp[i-1],j)
        dp.pop(0)
        print(*dp)

    except:
        pass
        
M = 998244353
P = 1000000007
 
if __name__ == '__main__':
    #for _ in range(I()):main()
    for _ in range(1):main()