Vasya 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 
#include <bits/stdc++.h>
using namespace std;
const int M = 5e6 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 9e18;
const double pi = 3.14159265358979323846;
long long int t, n, m, x, sum, k, p, u = 0;
string s, z, ans = "";
long long int P[10000005] = {0};
long long int raised(long long int x, long long int v) {
  if (v == 0) return 1 % mod;
  if (v % 2 == 1) return (x * raised(x, v - 1)) % mod;
  long long int r = raised(x, v / 2);
  return (r * r) % mod;
}
bool prime(long long int n) {
  if (n == 1) return 1;
  for (int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int hcf(long long int a, long long int b) {
  if (a % b == 0) return b;
  return hcf(b, a % b);
}
bool change(string a, string b) {
  int n = a.size();
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] || abs(a[i] - b[i]) == 32)
      continue;
    else
      return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n;
  for (int i = 1; i < n; i++) cout << i + 1 << " ";
  cout << "1";
}
