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.*;
import java.util.*;
import java.util.function.*;
import java.lang.*;

public class Main {
    final static String fileName = "";
    final static boolean useFiles = false;

    public static void main(String[] args) throws FileNotFoundException {
        InputStream inputStream;
        OutputStream outputStream;
        if (useFiles) {
            inputStream = new FileInputStream(fileName + ".in");
            outputStream = new FileOutputStream(fileName + ".out");
        } else {
            inputStream = System.in;
            outputStream = System.out;
        }
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task(in, out);
        Debug.out = out;
        solver.solve();
        out.close();
    }
}
class Task {
    public void solve() {
        int n = in.nextInt();
        int[] a = new int[2 * n - 1];
        for (int i = 0; i < n; i++)
            a[i] = i + 1;
        int head = 0;
        for (int step = 2; step <= n; step++) {
            for (int i = head + step; i <= head + n; i += Math.min(step, head + n - i)) {
                int t = a[head];
                a[head] = a[i];
                a[i] = t;
                if (i == head + n)
                    break;
            }
            head++;
        }
        for (int i = head; i < head + n; i++){
            out.print(a[i] + " ");
        }
    }

    private InputReader in;
    private PrintWriter out;

    Task(InputReader in, PrintWriter out) {
        this.in = in;
        this.out = out;
    }
}

class Debug {
    public static PrintWriter out;

    public static void printObjects(Object... objects) {
        out.println(Arrays.deepToString(objects));
        out.flush();
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

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

    public double nextDouble(){
        return Double.parseDouble(next());
    }

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

    public long nextLong(){
        return Long.parseLong(next());
    }

    public byte nextByte(){
        return Byte.parseByte(next());
    }
}