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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;

import static java.lang.Integer.*;
@SuppressWarnings("unused")

public class round176D {

    public static void main(String[] args)throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = parseInt(br.readLine());
        int [] ans = new int [2 * n];
        for(int i = 0 ; i < n ; ++i)
            ans[i] = i + 1;
        for(int i = 2 ; i <= n ; ++i){          
            int last = 0;
            int j;
            for(j = i - 2 ; j < (i - 2) + n ; j += i){
                int tmp = ans[j];
                ans[j] = last;
                last = tmp;
            }
            ans[i - 2 + n] = last;          
        }
        PrintWriter out = new PrintWriter(System.out);
        for(int i = n - 1 ; i < 2 * n - 1 ; ++i)
            out.print(ans[i] + " ");
        out.flush();
    }
}
