John Doe has found the beautiful permutation formula.

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

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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.PrintWriter;
import java.util.*;

public class Shifting {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		a = new int[3*n+1];
		for(int i=1; i<=n; i++) a[i] = i;
		for(int i=2; i<=n; i++){
			shift(i);
		}
		PrintWriter pw = new PrintWriter(System.out);
		for(int i=1; i<=n; i++) pw.print(a[cur+i-1]+" ");
		pw.println();
		pw.flush();
	}
	
	
	static void shift(int k){
		int r = (n-1)/k;
		a[cur+n] = a[k*r+cur];
		for(int i=k*(r-1)+cur; i>=cur; i-=k){
			a[i+k] = a[i];
		}
		cur++;
	}
	
	static int n;
	
	static int[] a;
	
	static int cur = 1;
	
}
