John Doe has found the beautiful permutation formula.

Let's take permutation p = p1, p2, ..., pn. Let's define transformation f of this permutation: 

<image>

where k (k > 1) is an integer, the transformation parameter, r is such maximum integer that rk ≤ n. If rk = n, then elements prk + 1, prk + 2 and so on are omitted. In other words, the described transformation of permutation p cyclically shifts to the left each consecutive block of length k and the last block with the length equal to the remainder after dividing n by k. 

John Doe thinks that permutation f(f( ... f(p = [1, 2, ..., n], 2) ... , n - 1), n) is beautiful. Unfortunately, he cannot quickly find the beautiful permutation he's interested in. That's why he asked you to help him.

Your task is to find a beautiful permutation for the given n. For clarifications, see the notes to the third sample.

Input

A single line contains integer n (2 ≤ n ≤ 106).

Output

Print n distinct space-separated integers from 1 to n — a beautiful permutation of size n.

Examples

Input

2


Output

2 1 


Input

3


Output

1 3 2 


Input

4


Output

4 2 3 1 

Note

A note to the third test sample: 

  * f([1, 2, 3, 4], 2) = [2, 1, 4, 3]
  * f([2, 1, 4, 3], 3) = [1, 4, 2, 3]
  * f([1, 4, 2, 3], 4) = [4, 2, 3, 1]
import java.io.*;
import java.util.*;
public class z4 { public static void main(String[] args) throws IOException {
	Scanner in = new Scanner(System.in);
	PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	int n=in.nextInt();
	int[] a=new int[n];
	for (int i=0;i<n;i++) a[i]=i+1;
	int nach=0;
	int temp;
	int posl;
	for (int i=2;i<=n;i++)
	{	
	  posl=(n-1)-(n-1)%i;
	  temp=a[(nach+posl)%n];
	  for (int j=posl;j>=i;j-=i) a[(nach+j)%n]=a[(nach+j-i)%n];
	  a[nach%n]=temp;
	  nach+=1;
	}
	for (int i=0;i<n;i++) {out.print(a[(i+nach)%n]);out.print(' ');}
	out.flush();
}
}