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 static java.util.Arrays.deepToString;

import java.io.*;
import java.math.*;
import java.util.*;

public class B {

	static void solve() {
		int n = nextInt();
		setTime();
		int[] p = new int[n];
		for (int i = 0; i < n; i++) {
			p[i] = i;
		}
		int shift = 0;
		for (int k = 2; k <= n; k++) {
			int tmp = p[shift];
			for (int i = 0; (i + 1) * k < n; i++) {
				int pos = ((i + 1) * k + shift) % n;
				int tmp2 = p[pos];
				p[pos] = tmp;
				tmp = tmp2;
			}
			p[shift] = tmp;
			shift++;
		}
		print(p, shift);
	}

	static void print(int[] p, int shift) {
		for (int i = 0; i < p.length; i++) {
			writer.print((p[(i + shift) % p.length] + 1) + " ");
		}
	}

	public static void main(String[] args) throws Exception {
		reader = new BufferedReader(new InputStreamReader(System.in));
		writer = new PrintWriter(System.out);
		Locale.setDefault(Locale.US);

		setTime();
		solve();
		printTime();
		printMemory();

		writer.close();
	}

	static BufferedReader reader;
	static PrintWriter writer;
	static StringTokenizer tok = new StringTokenizer("");
	static long systemTime;

	static void debug(Object... o) {
		System.err.println(deepToString(o));
	}

	static void setTime() {
		systemTime = System.currentTimeMillis();
	}

	static void printTime() {
		System.err.println("Time consumed: " + (System.currentTimeMillis() - systemTime));
	}

	static void printMemory() {
		System.err.println("Memory consumed: "
				+ (Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory()) / 1000 + "kb");
	}

	static String next() {
		while (!tok.hasMoreTokens()) {
			String w = null;
			try {
				w = reader.readLine();
			} catch (Exception e) {
				e.printStackTrace();
			}
			if (w == null)
				return null;
			tok = new StringTokenizer(w);
		}
		return tok.nextToken();
	}

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

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

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

	static BigInteger nextBigInteger() {
		return new BigInteger(next());
	}
}