Problem: An exam for n students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (i and i + 1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.

Your task is to choose the maximum number of students and make such an arrangement of students in the room that no two students with adjacent numbers sit side by side.

Input

A single line contains integer n (1 ≤ n ≤ 5000) — the number of students at an exam.

Output

In the first line print integer k — the maximum number of students who can be seated so that no two students with adjacent numbers sit next to each other.

In the second line print k distinct integers a1, a2, ..., ak (1 ≤ ai ≤ n), where ai is the number of the student on the i-th position. The students on adjacent positions mustn't have adjacent numbers. Formally, the following should be true: |ai - ai + 1| ≠ 1 for all i from 1 to k - 1.

If there are several possible answers, output any of them.

Examples

Input

6

Output

6
1 5 3 6 2 4

Input

3


Output

2
1 3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
vector<long long> vv;
set<long long> st;
long long x, y, z;
string s, s1, s2, s3, s4, s5;
void memset_ll(long long a[], long long v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
void memset_s(char a[], char v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
int compare(const void* a, const void* b) {
  return (*(long long*)a - *(long long*)b);
}
void q_sort(long long a[], long long n) {
  qsort(a, n, sizeof(long long), compare);
}
static int compare1(const void* a, const void* b) {
  if (*(double*)a > *(double*)b)
    return 1;
  else if (*(double*)a < *(double*)b)
    return -1;
  else
    return 0;
}
void q_sort_d(double a[], long long n) {
  qsort(a, n, sizeof(double), compare1);
}
string trans_upper(string ss) {
  transform(ss.begin(), ss.end(), ss.begin(), ::toupper);
  return ss;
}
long long power(long long a, long long b) {
  long long pow = 1;
  while (b) {
    if (b & 1) {
      pow = pow * a;
      --b;
    }
    a = a * a;
    b = b / 2;
  }
  return pow;
}
long long bsearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return bsearch(arr, l, mid - 1, x);
    return bsearch(arr, mid + 1, r, x);
  }
  return -1;
}
long long fact(long long n) {
  long long f = 1, i;
  for (i = 1; i <= n; i++) f = f * i;
  return f;
}
long long is_prime(long long n) {
  long long i, j, k = 1;
  for (i = 2; i <= n / 2; i++) {
    if (n % i == 0) {
      k = 0;
      break;
    }
  }
  return k;
}
long long vow(char c) {
  long long k = 1;
  long long l = 0;
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
    return k;
  else
    return l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, l, p, q, n, m, o = 0, r, i, j, h, k, e = 0, d = 0, f = 0, g = 0;
  long long temp = 0, nl, np;
  double p1, p2 = 0.0, p3 = 0.0, p4, p5, p6, pp = 0.0, ppp;
  p1 = 3.1415926536;
  long long x1, x2, y1, y2, uu = 0, dd = 0, rr = 0, lll = 0;
  cin >> n;
  if (n == 2) {
    cout << 1 << endl;
    cout << 1;
    return 0;
  }
  if (n == 3) {
    cout << 2 << endl;
    cout << 1 << " " << 3;
    return 0;
  }
  for (i = 1; i <= n; i += 2) v.push_back(i);
  for (i = 2; i <= n; i += 2) vv.push_back(i);
  l = v.size();
  t = vv.size();
  cout << n << endl;
  for (i = l - 1; i >= 0; i--) cout << v[i] << " ";
  for (i = t - 1; i >= 0; i--) cout << vv[i] << " ";
  return 0;
}