The Little Elephant enjoys recursive functions.

This time he enjoys the sorting function. Let a is a permutation of an integers from 1 to n, inclusive, and ai denotes the i-th element of the permutation. The Little Elephant's recursive function f(x), that sorts the first x permutation's elements, works as follows:

  * If x = 1, exit the function. 
  * Otherwise, call f(x - 1), and then make swap(ax - 1, ax) (swap the x-th and (x - 1)-th elements of a). 



The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 to n, such that after performing the Little Elephant's function (that is call f(n)), the permutation will be sorted in ascending order.

Input

A single line contains integer n (1 ≤ n ≤ 1000) — the size of permutation.

Output

In a single line print n distinct integers from 1 to n — the required permutation. Numbers in a line should be separated by spaces.

It is guaranteed that the answer exists.

Examples

Input

1


Output

1 

Input

2


Output

2 1 
import java.util.Scanner;

public class A_221_Little_Elephant_And_Function {
	
	public static void f(int x,int[] numbers){
		if(x==1){
			//do nothing
		}
		if(x>=2){
			int tmp;
			tmp=numbers[x];
			numbers[x]=numbers[x-1];
			numbers[x-1]=tmp;
			f(x-1,numbers);
		}
	}
	
	public static void main(String[] args){
		Scanner input=new Scanner(System.in);
		int n=input.nextInt();
		int[] numbers=new int[n+1];
		int i;
		for(i=1;i<=n;i++)
			numbers[i]=i;
		f(n,numbers);
		for(i=1;i<=n;i++)
			System.out.print(numbers[i]+" ");
	}
}
