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.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author ocelopilli
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskD {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int[] p = new int[ 2*n + 1 ];

        for (int i=0; i<n; i++) p[i+1] = i+1;

        int begin = 1;
        for (int k=2; k<=n; k++)
        {
            int aux = p[ begin ];
            for (int i = 0; k * i < n; i++)
            {
                aux ^= p[ begin + k * i ];
                p[ begin + k * i ] ^= aux;
                aux ^= p[ begin + k * i ];
            }
            p[ begin + n ] = aux;
            begin++;
        }

        for (int i=0; i<n; i++) out.print( p[begin + i]+" " );
    }
}

class InputReader {
    BufferedReader br;
    StringTokenizer st;

    public InputReader(InputStream in)
    {
        br = new BufferedReader(new InputStreamReader(in));
        st = null;
    }

    public String next()
    {
        while (st==null || !st.hasMoreTokens())
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }

    public int nextInt() { return Integer.parseInt(next()); }
    }
