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:
from collections import Counter
import string
import math
import bisect
#import random
import sys
# sys.setrecursionlimit(10**6) 
from fractions import Fraction
def array_int():
    return [int(i) for i in sys.stdin.readline().split()]
def vary(arrber_of_variables):
    if arrber_of_variables==1:
        return int(sys.stdin.readline())
    if arrber_of_variables>=2:
        return map(int,sys.stdin.readline().split()) 
def makedict(var):
    return dict(Counter(var))
def primeFactors(n): 
          
        # Print the number of two's that divide n 
        primes=[]
        while n % 2 == 0: 
            primes.append(2) 
            n = n // 2
              
        # n must be odd at this point 
        # so a skip of 2 ( i = i + 2) can be used 
        for i in range(3,int(math.sqrt(n))+1,2): 
              
            # while i divides n , print i ad divide n 
            while n % i== 0: 
                primes.append(i) 
                n = n // i 
                  
        # Condition if n is a prime 
        # number greater than 2 
        if n > 2: 
            primes.append(n)
        return primes
primes=[]
def SieveOfEratosthenes(n):
    global primes
    # 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 * p, n+1, p):
                prime[i] = False
        p += 1
 
    # Print all prime numbers
    return prime
prime=SieveOfEratosthenes(10**5+1)

testcases=1
for _ in range(testcases):
    n=vary(1)
    num=[1]*(n+1)
    j=2
    for i in range(3,n+1):
        if prime[i]:
            num[i]=j
            j+=1
        else:
            # print(prime[i])
            primes=primeFactors(i)
            # print(primes)
            primes.sort()
            num[i]=num[primes[0]]
    print(*(num[2:]))