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.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;

public class ProblemB {

	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();

		int[] a = new int[2*n];
		for (int i = 0 ; i < n ; i++) {
			a[i] = i+1;
		}
		
		int head = 0;
		for (int p = 2 ; p <= n ; p++) {
			int t = head;
			int end = head + n;
			int temp = a[t];
			while (t < end) {
				int go = Math.min(end, t + p);
				int te = a[go];
				a[go] = temp;
				temp = te;
				t = go;
			}
			head++;
		}
		
		StringBuffer b = new StringBuffer();
		for (int i = n - 1 ; i < 2 * n - 1 ; i++) {
			b.append(" ").append(a[i]);
		}
		out.println(b.substring(1));
		out.flush();
	}
	

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