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.*;

public class LittleElephantFunction {
	static Scanner scn = new Scanner(System.in);

	public static void main(String[] args) {
		helper();
	}

	public static void helper() {

		int n = (int)input();
		print(n + " ");
		for (int i = 1; i < n; i++)
			print(i + " ");

		print();

	}

	public static long input() {
		return scn.nextLong();
	}

	public static String input(int... var) {
		return scn.next();
	}

	public static void TakeInput(int[] arr) {
		for (int i = 0; i < arr.length; i++)
			arr[i] = (int)input();
	}

	public static <T> void printArray(T[] arr) {
		for (T val : arr)
			print(val + " ");

		print("\n");
	}

	public static <T> void print(T t) {
		System.out.print(t);
	}
	public static void print() {
		System.out.println();
	}
}