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 sys,math,io,os,time,itertools,collections
mod=10**9+7
sys.setrecursionlimit(10000)
i=sys.stdin.readline
#use sys.stdout.write() (remember to convert to str b4 and concatenate "\n")
global start,end

#sieve of eratosthenes
def sieve(n):
    prime=[0 for k in range(n+1)]
    p=2
    color=1
    while p<=n:
        if prime[p]==0:
            prime[p]=color
            for k in range(p*p,n+1,p):
                prime[k]=color
            color+=1
        p+=1
    return prime[2:]

def main():
    n=int(i())
    print(*sieve(n))

t=1
#t=int(i())
start=time.perf_counter()
for _ in range(t):
    main()
end=time.perf_counter()
#print(end-start)