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 sieve(n):
    arr = [1]*(n+1)
    arr[0], arr[1] = 0,0
    
    for i in range(2, n+1):
        if arr[i] == 1:
            arr[2*i::i] = [0 for _ in range(len(arr[2*i::i]))]
            
    return arr

def min_prime_factors(n):
    factors=[]
    d=2
    while(d*d<=n):
        while(n>1):            
            while n%d==0:
                factors.append(d)
                n=n/d
            d+=1
    return factors[-1]

def solve():
    N = int(input())
    
    prime = sieve(N+1)

    result = [1]*(N+1)
    result[3::2] = [2 for _ in range(len(result[3::2]))]

    last = 3
    
    for i in range(5, N+1, 2):
        if prime[i]:
            result[i::i] = [last for _ in range(len(result[i::i]))]
            last += 1

    print (' '.join(map(str, result[2:])))
    
if __name__ == "__main__":  
    solve()