Package for this problem was not updated by the problem writer or Codeforces administration after we’ve upgraded the judging servers. To adjust the time limit constraint, solution execution time will be multiplied by 2. For example, if your solution works for 400 ms on judging servers, then value 800 ms will be displayed and used to determine the verdict.

Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

The problem statement has recently been changed. View the changes.

×
C. Disposition

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputVasya bought the collected works of a well-known Berland poet Petya in *n* volumes. The volumes are numbered from 1 to *n*. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers *i* such that for at least one *j* (1 ≤ *j* ≤ *n*) is true both: *j* *mod* *i* = 0 and at the same time *p*(*j*) *mod* *i* = 0, where *p*(*j*) is the number of the tome that stands on the *j*-th place and *mod* is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.

Help Vasya — find the volume disposition with the minimum number of divisors.

Input

The first line contains number *n* (1 ≤ *n* ≤ 100000) which represents the number of volumes and free places.

Output

Print *n* numbers — the sought disposition with the minimum divisor number. The *j*-th number (1 ≤ *j* ≤ *n*) should be equal to *p*(*j*) — the number of tome that stands on the *j*-th place. If there are several solutions, print any of them.

Examples

Input

2

Output

2 1

Input

3

Output

1 3 2

Codeforces (c) Copyright 2010-2021 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: May/13/2021 08:25:33 (h2).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|