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.*;
import java.io.*;

public class MainApp
{
    public static void main(String[] args) throws IOException {
        FastScanner fs = new FastScanner();
        int n = fs.nextInt();
        System.out.print(n+" ");
        for (int i = 1; i < n; i++) {
            System.out.print(i+" ");
        }
    }
}

class FastScanner
{
    private BufferedReader bf;
    private StringTokenizer st;

    public FastScanner()
    {
        bf = new BufferedReader(new InputStreamReader(System.in));
        st = null;
    }

    public String next()
    {
        while (st == null || !st.hasMoreTokens())
        {
            try
            {
                st = new StringTokenizer(bf.readLine());
            }
            catch (IOException e)
            {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }

    public int nextInt()
    {
        return Integer.parseInt(this.next());
    }
    public long nextLong() { return Long.parseLong(this.next()); }
}
