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:
def primen(n):
    for i in range(2,n):
        if n % i == 0:
            return i

n = int(input())
idxs = []
for i in range(n+2):
    idxs.append(i+1)
if n == 1:
    print(1)
    exit()
res = [0]*(n-1)
prime = [True for i in range(100002)] 
p = 2
arr = []
count = 0
while (p * p <= 100002): 
    if (prime[p] == True): 
        for i in range(p * p, 100002, p): 
            prime[i] = False
    p += 1
for p in range(2, 100002): 
    if prime[p]: 
        arr.append(p)
nexti = arr[0]
together = dict(zip(arr,idxs))
for i in range(n-1):
    if (i+2) == nexti:
        idx = together.get(i+2)
        res[i] = idx
        count += 1
        if count >= len(arr):
            continue
        nexti = arr[count] 
    else:
        no = primen(i+2)
        idx = together.get(no)
        res[i] = idx
print(*res)