Problem: Given an integer N, find two permutations:

  1. Permutation p of numbers from 1 to N such that pi ≠ i and pi & i = 0 for all i = 1, 2, ..., N. 
  2. Permutation q of numbers from 1 to N such that qi ≠ i and qi & i ≠ 0 for all i = 1, 2, ..., N. 



& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

Input

The input consists of one line containing a single integer N (1 ≤ N ≤ 105).

Output

For each subtask, if the required permutation doesn't exist, output a single line containing the word "NO"; otherwise output the word "YES" in the first line and N elements of the permutation, separated by spaces, in the second line. If there are several possible permutations in a subtask, output any of them.

Examples

Input

3


Output

NO
NO


Input

6


Output

YES
6 5 4 3 2 1 
YES
3 6 2 5 1 4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int SIZE = 1 << 10;
int pointer = SIZE;
char buffer[SIZE];
char Advance() {
  if (pointer == SIZE) {
    fread(buffer, 1, SIZE, stdin);
    pointer = 0;
  }
  return buffer[pointer++];
}
int Read() {
  int answer = 0;
  char ch = Advance();
  while (!isdigit(ch)) ch = Advance();
  while (isdigit(ch)) {
    answer = answer * 10 + ch - '0';
    ch = Advance();
  }
  return answer;
}
const int MAXN = 100000;
int answer[1 + MAXN];
int main() {
  int n;
  scanf("%d", &n);
  int initial = n;
  if (n % 2)
    printf("NO\n");
  else {
    printf("YES\n");
    while (n) {
      int k = 0;
      while ((1 << k) <= n) k++;
      k--;
      int a = (1 << k) - 1, b = (1 << k);
      while (b <= n) {
        answer[a] = b;
        answer[b] = a;
        a--;
        b++;
      }
      n = a;
    }
    n = initial;
    for (int i = 1; i <= n; i++) printf("%d ", answer[i]);
    printf("\n");
  }
  if (n <= 5 || n == (n & -n))
    printf("NO\n");
  else {
    printf("YES\n");
    if (n == 6)
      printf("3 6 2 5 1 4");
    else {
      printf("7 3 6 5 1 2 4");
      int a = 8;
      while (a <= n) {
        int b = min(n, 2 * a - 1);
        for (int i = a; i < b; i++) printf(" %d", i + 1);
        printf(" %d", a);
        a = b + 1;
      }
    }
  }
  return 0;
}