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;
int N;
int Ans[100001];
int Log[100001];
bool Search(int Num) {
  if (Num == 1) return false;
  if (!Num) return true;
  int Total = (1 << (Log[Num] + 1)) - 1;
  if (!Search(Total - Num - 1)) return false;
  for (int i = Total - Num; i <= Num; ++i) Ans[i] = Total - i;
  return true;
}
int main() {
  scanf("%d", &N);
  Log[0] = -1;
  for (int i = 1; i <= N; ++i) Log[i] = Log[i >> 1] + 1;
  if (Search(N)) {
    printf("YES\n");
    for (int i = 1; i <= N; ++i) printf("%d ", Ans[i]);
    printf("\n");
  } else
    printf("NO\n");
  if (N <= 5)
    printf("NO\n");
  else if (N == 6)
    printf("YES\n5 3 2 6 1 4\n");
  else {
    if ((1 << Log[N]) == N) {
      printf("NO\n");
      return 0;
    }
    Ans[1] = 7, Ans[2] = 3, Ans[3] = 2, Ans[4] = 5, Ans[5] = 6, Ans[6] = 4,
    Ans[7] = 1;
    int Pos = 8;
    while (Pos <= N) {
      int i;
      for (i = Pos; Log[i + 1] == Log[Pos] && i <= N; ++i) Ans[i] = i + 1;
      Ans[i] = Pos;
      Pos = i + 1;
    }
    printf("YES\n");
    for (int i = 1; i <= N; ++i) printf("%d ", Ans[i]);
    printf("\n");
  }
  return 0;
}