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.*;

public class Main {

	public static void main(String[] args) throws Exception {
		long time = System.nanoTime();

		int n = next();

		int len = n;
		for (int i = 2; i <= n; i++) len += (n + i - 1)/i;

//		out.println(len);

		int[] x = new int[len + 5];
		for (int i = 0; i < n; i++) x[i] = i;
		int start = 0;
		for (int i = 2; i <= n; i++, start++) {
			int j = i * (n/i) + start;
			x[start + n] = x[j];
			for (; j > start; j -= i) x[j] = x[j - i];
		}

		for (int i = start; i < start + n; i++) out.print(++x[i] + " ");
		out.println();


		
//		System.out.println((System.nanoTime() - time)* 1e-9);
		out.close();
	}
	static PrintWriter out = new PrintWriter(System.out);
	
	static BufferedReader bufferedreader = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer in = new StringTokenizer("");

	static String nextToken() throws Exception {
		if (!in.hasMoreTokens()) in = new StringTokenizer(bufferedreader.readLine());
		return in.nextToken();
	}

	static int next()  throws Exception {return Integer.parseInt(nextToken());};
	static int[] next(int n) throws Exception {
		int[] x = new int[n];
		for (int i = 0; i < n; i++) x[i] = next();
		return x;
	}
	static int[][] next(int n, int m) throws Exception {
		int[][] x = new int[n][];
		for (int i = 0; i < n; i++) x[i] = next(m);
		return x;
	}

	static long nextl() throws Exception {return Long.parseLong(nextToken());};
	static long[] nextl(int n) throws Exception {
		long[] x = new long[n];
		for (int i = 0; i < n; i++) x[i] = nextl();
		return x;
	}
	static long[][] nextl(int n, int m) throws Exception {
		long[][] x = new long[n][];
		for (int i = 0; i < n; i++) x[i] = nextl(m);
		return x;
	}

	static double nextd() throws Exception {return Double.parseDouble(nextToken());};
	static double[] nextd(int n) throws Exception {
		double[] x = new double[n];
		for (int i = 0; i < n; i++) x[i] = nextd();
		return x;
	}
	static double[][] nextd(int n, int m) throws Exception {
		double[][] x = new double[n][];
		for (int i = 0; i < n; i++) x[i] = nextd(m);
		return x;
	}

	static String nextline() throws Exception {
		in = new StringTokenizer("");
		return bufferedreader.readLine();
	}

}