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:
primes=[]
def SieveOfEratosthenes(n): 
    prime=[True for i in xrange(n+1)] 
    p=2
    while (p*p<=n):           
        if (prime[p]==True): 
            for i in xrange(p*2,n+1,p): 
                prime[i]=False
        p+=1
    for p in xrange(2,n): 
        if prime[p]: 
            primes.append(p)
        else:
            primes.append("#")
n=10**6
SieveOfEratosthenes(n)
primes=["#","#"]+primes

import math
n=int(input())
count=7
l=[".","."]
for i in xrange(2,n+1):
    if i%2==0:
        l.append(1)
    elif i%3==0:
        l.append(2)
    elif i%5==0:
        l.append(3)
    elif i%7==0:
        l.append(4)
    elif i%11==0:
        l.append(5)
    elif i%13==0:
        l.append(6)
    else:
        if primes[i]!="#":
            l.append(count)
            count+=1
        elif primes[i]=="#":
            for j in xrange(2,int(math.sqrt(i))+1):
                if i%j==0:
                    l.append(l[j])
                    break
for i in l:
    if i!=".":
        print i,