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.Arrays;
import java.util.Scanner;

/**
 * Created: 2/27/2018
 *
 * @author phil
 */
public class A221_LittleElephantAndFunction {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int a[] = new int[n];

		for (int i=0; i<n; i++) {
			if (i == 0) {
				a[i] = n;
			} else {
				a[i] = i;
			}
		}

		for (int i: a) {
			System.out.print(i + " ");
		}


//		System.out.println(Arrays.toString(a));

//		elephant(n-1,a);
//		System.out.println(Arrays.toString(a));

	}

	public static void elephant(int x, int[] a) {
		if (x == 0) {
			return;
		} else {
			elephant(x-1, a);
			int temp = a[x];
			a[x] = a[x-1];
			a[x-1] = temp;
		}


	}

}
