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.math.*;

public class Main {
	public static void main(String[] args) {
		new Main().run();
	}
	void run() {
		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = in.nextInt();
		int[] pos = new int[n + n];
		for (int i = 0; i < n; ++i)
			pos[i] = i + 1;
		int count = 0;
		for (int i = 2; i <= n; ++i) {
			int rem = n % i;
			if (rem > 0)
				pos[n + i - 2] = pos[n / i * i + i - 2];
			for (int j = n / i * i + i - 2; j > i - 2; j -= i)
				pos[j] = pos[j - i];
		}

		for (int i = n - 1; i < n + n - 1; ++i)
			out.print(pos[i] + " ");
		out.println();

		out.close();
	}
}

class InputReader {
	BufferedReader buff;
	StringTokenizer tokenizer;

	InputReader(InputStream stream) {
		buff = new BufferedReader(new InputStreamReader(stream));
		tokenizer = null;
	}
	boolean hasNext() {
		while (tokenizer == null || !tokenizer.hasMoreTokens())
			try {
				tokenizer = new StringTokenizer(buff.readLine());
			}
			catch (Exception e) {
				return false;
			}
		return true;
	}
	String next() {
		if (!hasNext())
			throw new RuntimeException();
		return tokenizer.nextToken();
	}
	int nextInt() { return Integer.parseInt(next()); }
	long nextLong() { return Long.parseLong(next()); }
}

