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.*;
import java.lang.*;
import java.math.*;
import java.io.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class B{
	Scanner sc=new Scanner(System.in);

	int INF=1<<28;
	double EPS=1e-9;

	void test(){
		for(int n=7; n<=7; n++){
			int[] a=new int[n];
			int[] next=new int[n];
			for(int i=0; i<n; i++){
				a[i]=i+1;
			}
			for(int k=2; k<=n*0+10; k++){
				for(int offset=0; offset<n; offset+=k){
					int d=min(k, n-offset);
					for(int i=0; i<d; i++){
						next[offset+i]=a[offset+(i+1)%d];
					}
				}
				debug(next);
				a=next.clone();
			}
			String s="";
			for(int i=0; i<n; i++){
				s+="    ";
			}
			for(int i : a){
				s+=String.format("%2d, ", i);
			}
			debug(s);
		}
	}

	int n;

	void run(){
		// test();
		n=sc.nextInt();
		solve();

	}

	void solve(){
		int[] a=new int[2*n];
		for(int i=0; i<n; i++){
			a[i]=i+1;
		}
		int os=0;
//		debug(a);
		for(int d=2; d<=n; d++){
//			debug("d", d);
//			debug("last", (n-1)/d*d);
			for(int i=(n-1)/d*d; i+os>=0; i-=d){
				a[min(i+(d-1), n-1)+(os+1)]=a[i+os];
			}
			a[os]=0;
//			debug(a);
			os++;
		}
		StringBuilder sb=new StringBuilder();
		for(int i : a){
			if(i>0){
				sb.append(i).append(' ');
			}
		}
		println(sb.toString().trim());
	}

	void println(String s){
		System.out.println(s);
	}

	void print(String s){
		System.out.print(s);
	}

	void debug(Object... os){
		System.err.println(Arrays.deepToString(os));
	}

	public static void main(String[] args){
		Locale.setDefault(Locale.US);
		new B().run();
	}
}
