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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.logging.Level;
import java.util.logging.Logger;

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author pttrung
 */
public class Code {

    public static void main(String[] args) {
        try {
            BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

            int n = Integer.parseInt(in.readLine());
            System.out.print(n);
            for(int i = 1; i < n; i++){
                System.out.print(" " + i);
            }
            System.out.println();

        } catch (IOException ex) {
            Logger.getLogger(Code.class.getName()).log(Level.SEVERE, null, ex);
        }
    }

    public static int count(long val) {
        int count = 0;
        while (val > 0) {
            if (val % 10 == 9) {
                count++;
            } else {
                return count;
            }
            val /= 10;
        }
        return count;
    }
}
