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 LittleElephantAndFunc {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int a[]=new int[n];
        for (int i=1;i<=n;i++){
            a[i-1]=i;
        }
        a= sort(a, a.length-1);
        for(int i: a)
            System.out.print(i+" ");
        }
        public static int[] sort(int a[],int x) {
            if (x == 0) {
                return a;
            } else {
                int t = a[x];
                a[x] = a[x - 1];
                a[x-1]=t;
                return sort(a,x-1);
            }
        }
    }

