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.Scanner;

public class D {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] ar=new int[2*n];
        for(int i=0;i<n;i++){
            ar[i]=i+1;
        }
        int shift=0;
        for(int i=2;i<=n;i++){
            int start=(((int)Math.ceil(n/(1.0*i)))-1)*i;
            ar[n+shift]=ar[start+shift];
            for(int j=start-i;j>=0;j-=i){
                ar[j+i+shift]=ar[j+shift];
            }
            shift++;
        }
        PrintWriter out=new PrintWriter(System.out);
        for(int i=n-1,j=0;j<n;i++,j++){
            out.print(ar[i]+" ");
        }
        out.println();
        out.flush();
    }
}
