#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
inline bool isDigit(char c) { return '0' <= c && c <= '9'; }
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline bool isSquare(T x) {
  T y = sqrt(x + 0.5);
  return (y * y) == x;
}
template <class T1, class T2>
inline T1 gcd(T1 a, T2 b) {
  return b ? gcd(b, a % b) : a;
}
template <class T1, class T2>
inline T1 eqMin(T1 &x, const T2 &y) {
  if (T1(y) < x) return x = y;
  return x;
}
template <class T1, class T2>
inline T1 eqMax(T1 &x, const T2 &y) {
  if (T1(y) > x) return x = y;
  return x;
}
template <class T1, class T2>
inline T1 min(const T1 &x, const T2 &y) {
  return x < (T1)y ? x : (T1)y;
}
template <class T1, class T2>
inline T1 max(T1 &x, const T2 &y) {
  return x > (T1)y ? x : (T1)y;
}
template <typename T>
inline T getint() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * p;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T>
auto operator<<(ostream &os, const T &v)
    -> decltype(v.begin(), v.end(), declval<ostream &>()) {
  os << '{';
  bool was = false;
  for (const auto &x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << '}';
  return os;
}
template <size_t sz>
ostream &operator<<(ostream &os, const bitset<sz> &bit) {
  os << '/';
  for (int i = 0; i < sz; i++) os << bit[i];
  os << '\\';
  return os;
}
const double PI = acos(-1);
const double EPS = 1e-8;
const int INF = (int)2e9;
const long long LINF = (long long)2e18;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1;
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  deque<int> v;
  for (int i = 1; i <= n; i++) {
    v.push_back(i);
  }
  v.push_front(v.back());
  v.pop_back();
  for (int it : v) {
    cout << it << ' ';
  }
}

### Prompt

Your task is to create a CPP solution to the following problem:

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 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
inline bool isDigit(char c) { return '0' <= c && c <= '9'; }
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline bool isSquare(T x) {
  T y = sqrt(x + 0.5);
  return (y * y) == x;
}
template <class T1, class T2>
inline T1 gcd(T1 a, T2 b) {
  return b ? gcd(b, a % b) : a;
}
template <class T1, class T2>
inline T1 eqMin(T1 &x, const T2 &y) {
  if (T1(y) < x) return x = y;
  return x;
}
template <class T1, class T2>
inline T1 eqMax(T1 &x, const T2 &y) {
  if (T1(y) > x) return x = y;
  return x;
}
template <class T1, class T2>
inline T1 min(const T1 &x, const T2 &y) {
  return x < (T1)y ? x : (T1)y;
}
template <class T1, class T2>
inline T1 max(T1 &x, const T2 &y) {
  return x > (T1)y ? x : (T1)y;
}
template <typename T>
inline T getint() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * p;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T>
auto operator<<(ostream &os, const T &v)
    -> decltype(v.begin(), v.end(), declval<ostream &>()) {
  os << '{';
  bool was = false;
  for (const auto &x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << '}';
  return os;
}
template <size_t sz>
ostream &operator<<(ostream &os, const bitset<sz> &bit) {
  os << '/';
  for (int i = 0; i < sz; i++) os << bit[i];
  os << '\\';
  return os;
}
const double PI = acos(-1);
const double EPS = 1e-8;
const int INF = (int)2e9;
const long long LINF = (long long)2e18;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1;
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  deque<int> v;
  for (int i = 1; i <= n; i++) {
    v.push_back(i);
  }
  v.push_front(v.back());
  v.pop_back();
  for (int it : v) {
    cout << it << ' ';
  }
}

```