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.math.BigInteger;
import java.util.*;

public class A_Little_Elephant_and_Function {
    public static void main (String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        System.out.print(n + " ");
        for (int i=1; i<n; i++) {
            System.out.print(i + " ");
        }
        scan.close();
    }
    public static void inputArray (int[] arr, int n, Scanner scan) {
        for (int i=0; i<n; i++) {
            arr[i] = scan.nextInt();
        }
    }
    public static void printArray (int[] arr) {
        for (int i=0; i<arr.length; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }
    public static boolean isPrime (int n) {
        BigInteger a = BigInteger.valueOf(n);
        return a.isProbablePrime(1);
    }
    public static int gcd (int a, int b) {
        BigInteger a1 = BigInteger.valueOf(a);
        BigInteger b1 = BigInteger.valueOf(b);
        return a1.gcd(b1).intValue();
    }
    public static int numberOfDigitsInNumber (int n) {
        return (int)Math.floor(Math.log10(n)) + 1;
    }
    public static boolean isPowerOfTwo (int x) {
        return x!=0 && ((x&(x-1)) == 0);   
    }
    public static String toBinaryString (int n) {
        return Integer.toString(n, 2);
        // for other bases change 2 to other bases
    }
}