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 maxn = 1e5 + 10, mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int bin[20];
int ans[maxn];
int base7[10] = {0, 3, 6, 1, 5, 7, 2, 4};
int base6[10] = {
    0, 3, 6, 2, 5, 1, 4,
};
void solve1(int n) {
  int pp = n;
  if (n & 1) {
    printf("NO\n");
    return;
  }
  printf("YES\n");
  int t;
  while (n) {
    for (int i = 0; i < 20; i++)
      if (bin[i] <= n) t = i;
    for (int j = bin[t]; j <= n; j++)
      ans[j] = bin[t] * 2 - 1 - j, ans[bin[t] * 2 - 1 - j] = j;
    n -= 2 * (n - bin[t] + 1);
  }
  for (int i = 1; i <= pp; i++) printf("%d ", ans[i]);
  printf("\n");
}
void solve2(int n) {
  if (n <= 5) {
    printf("NO\n");
    return;
  }
  for (int i = 3; i < 20; i++)
    if (n == bin[i]) {
      printf("NO\n");
      return;
    }
  printf("YES\n");
  if (n == 6) {
    for (int i = 1; i <= 6; i++) printf("%d ", base6[i]);
    return;
  }
  if (n == 7) {
    for (int i = 1; i <= 7; i++) printf("%d ", base7[i]);
    return;
  }
  int t;
  for (int i = 0; i < 20; i++)
    if (bin[i] < n) t = i;
  for (int i = 1; i <= 7; i++) ans[i] = base7[i];
  for (int i = 3; i < t; i++) {
    for (int j = bin[i]; j < bin[i + 1]; j++) {
      if (j == bin[i + 1] - 1)
        ans[j] = bin[i];
      else
        ans[j] = j + 1;
    }
  }
  for (int i = bin[t]; i <= n; i++) {
    if (i == n)
      ans[i] = bin[t];
    else
      ans[i] = i + 1;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
int main() {
  for (int i = 0; i < 20; i++) bin[i] = 1 << i;
  int n;
  scanf("%d", &n);
  solve1(n);
  solve2(n);
  return 0;
}