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:
import math as mt   
MAXN = 100001
spf = [0 for i in range(MAXN)] 
   
def sieve(): 
    spf[1] = 1
    for i in range(2, MAXN):  
        spf[i] = i 
    for i in range(4, MAXN, 2): 
        spf[i] = 2
  
    for i in range(3, mt.ceil(mt.sqrt(MAXN))):  
        if (spf[i] == i):
            for j in range(i * i, MAXN, i):  
                if (spf[j] == j): 
                    spf[j] = i 
  
def getFactorization(x): 
    ret = list() 
    while (x != 1): 
        ret.append(spf[x]) 
        x = x // spf[x] 
  
    return ret 

sieve()
n = int(input())
seq = []
color = [0] * (n+1)
i = 1
for e in range(2, n + 1):
    seq.append(e)
for c in range(len(seq)):
    num = seq[c]
    mp = getFactorization(num)[0]
    if(color[mp] == 0):
        color[mp] = i
        i+=1
    seq[c] = color[mp]
for w in range(len(seq)):
    print(seq[w], end=' ')