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.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author babkin_sv
 */
public class Solver {
    
    public static void main(String[] args) throws FileNotFoundException, IOException{
        new Solver().Run();
    }
    
    StringTokenizer Stok;
    BufferedReader br;
    PrintWriter pw;
    
    String nextToken() throws IOException{
        while (Stok==null || !Stok.hasMoreTokens()){
            Stok=new StringTokenizer(br.readLine());
        }
        return Stok.nextToken();
    }
    
    int nextInt() throws IOException{
        return Integer.parseInt(nextToken());
    }
    
    long nextLong() throws IOException{
        return Long.parseLong(nextToken());
    }
    
    double nextDouble() throws IOException{
        return Double.parseDouble(nextToken());
    }
    
    int kolElem;
    
    void Run() throws FileNotFoundException, IOException{
        //br=new BufferedReader(new FileReader("input.txt"));
        //pw=new PrintWriter("output.txt");
        br=new BufferedReader(new InputStreamReader(System.in));
        pw=new PrintWriter(new OutputStreamWriter(System.out));
        kolElem=nextInt();
        
        int[] elems=new int[2*kolElem];
        for (int i=0; i<kolElem; i++){
            elems[i]=i+1;
        }
        int nach=0;
        int kon=kolElem;
        
        for (int z=2; z<=kolElem; z++){
            int curInd=nach;
            int nextInd;
            int buf=elems[curInd]; //предыдущий элемент, который надо вставить на новое место
            int buf2;  //следующий элемент, который надо заменить
            while (curInd<kon){
                nextInd=curInd+z;
                if (nextInd<kon){
                    buf2=elems[nextInd];
                    elems[nextInd]=buf;
                    buf=buf2;
                } else {
                    nextInd=kon;
                    elems[nextInd]=buf;
                }
                curInd=nextInd;
            }
            nach++;
            kon++;
        }
        
        for (int i=nach; i<kon; i++){
            pw.print(elems[i]);
            pw.print(' ');
        }
        pw.println();
        pw.flush();
        pw.close();
    }
    
}
