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.Scanner;


public class B {
	public static final boolean DEBUG = false;
	Scanner sc;
	
	public void debug(Object o) {
		if (DEBUG) {
			int ln = Thread.currentThread().getStackTrace()[2].getLineNumber();
			String fn = Thread.currentThread().getStackTrace()[2].getFileName();
			System.out.println("(" + fn + ":" + ln+ "): " + o);
		}
	}
	
	public void pln(Object o) {
		System.out.println(o);
	}
	public void print(Object o) {
		System.out.print(o);
	}
	
	
	public void run() {
		sc = new Scanner(System.in);		
		int n = sc.nextInt();
		
		int[] p = new int[2000010];
		int st = 0;
		for (int i=1; i<=n; i++) {
			p[i] = i;
		}
		
		for (int i=2; i<=n; i++) {
			st = st + 1;	// left shift
			int j;
			
			int a = n/i * i;
			p[st+n] = p[st + a];
			for (j=st+a; j>=st+1; j-=i) {
				p[j] = p[j-i];
			}

//			for (int i2=st+1; i2<=st+n; i2++) {
//				print(p[i2] + " ");
//			}
//			pln("");
			
		}
		StringBuilder sb = new StringBuilder();
		for (int i=st+1; i<=st+n; i++) {
			sb.append(p[i] + " ");
		}
		System.out.println(sb);
	}
	public static void main(String[] args) {
		B t = new B();
		t.run();
	}
}
