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.util.*;
public class Main
{
	public static void main(String[] args)
	{
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] ary = new int[n * 2 + 5];
		for (int i = 1; i <= n; i++)
			ary[i] = i;
		for (int step = 1; step <= n - 1; step ++)
		{
			int k = step + 1;
			int head = step;
			int last = step + n - 1;
			int prev = 0;
			for (int j = head; j <= last; j += k)
			{
				int tmp = ary[j];
				ary[j] = prev;
				prev = tmp;
			}
			ary[last + 1] = prev;
		}
		StringBuilder sb = new StringBuilder();
		for (int i = 1; i <= n; i++)
		{
			sb.append(ary[i + n - 1]);
			if (i + 1 <= n)
			{
				sb.append(" ");
			}
		}
		System.out.println(sb);
	}
}
