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:
'''input
15
'''
# its just a simulation
from sys import stdin
import heapq
import collections

def get_spf():
	mx = 10 ** 5
	spf = [x for x in  range(mx + 1)]
	for i in range(2, mx + 1):
		if spf[i] == i:
			for j in range(i * i, mx + 1, i):
				if spf[j] == j:
					spf[j] = i

	prime = dict()
	for i in range(2, mx + 1):
		if spf[i] == i:
			prime[i] = True
	return spf, prime


# main starts
n = int(stdin.readline().strip())
spf, prime = get_spf()
arr = [1] * (n + 1)
max_prime = 2
arr[2] = 1
for i in range(3, n + 1, 2):
	if i == spf[i]:
		arr[i] = max_prime
		max_prime += 1
	else:
		arr[i] = arr[spf[i]]
print(*arr[2:])