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:
check = []
def SieveOfEratosthenes(n): 
      
    # 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
      
    
    for p in range(2, n): 
        if prime[p]: 
            check.append(p) 
#SieveOfEratosthenes(300000)
n=int(input())
prime = [True for i in range(n+1)] 
ans=[1 for i in range(n+1)]
count=1
p=2
while p<=n:
    #print (p)
    if prime[p]==True:
        ans[p]=count
        #print (ans[i])
        for i in range(p*p,n+1,p):
            prime[i]=False
            ans[i]=count
        count += 1
    p += 1
#print(ans)
#print(prime)
for i in range(2,n+1):
    print(ans[i],end=" ")