#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int ans[N];
void solve(int n) {
  if (n == 0) return;
  for (int j = 20; j >= 0; j--) {
    if ((1 << j) > n) continue;
    int cen = (1 << j);
    int ada = n - cen + 1;
    for (int i = 0; i < ada; i++) {
      swap(ans[cen + i], ans[cen - i - 1]);
    }
    return solve(cen - ada - 1);
  }
  assert(0);
}
void solve_1(int n) {
  if (n % 2) {
    puts("NO");
    return;
  }
  puts("YES");
  for (int i = 1; i <= n; i++) ans[i] = i;
  solve(n);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
}
void solve_2(int n) {
  if (n <= 5) {
    puts("NO");
    return;
  }
  if (__builtin_popcount(n) == 1) {
    puts("NO");
    return;
  }
  puts("YES");
  if (n == 6) {
    puts("3 6 1 5 4 2");
    return;
  }
  if (n == 7) {
    puts("3 6 1 5 4 7 2");
    return;
  }
  for (int i = 1; i <= n; i++) ans[i] = i;
  for (int j = 17; j >= 3; j--) {
    if ((1 << j) > n) continue;
    int from = (1 << j);
    int to = min(n, from * 2 - 1);
    for (int i = from; i <= to; i++) {
      if (i + 2 <= to) {
        swap(ans[i], ans[i + 2]);
      } else if (i + 1 <= to) {
        swap(ans[i], ans[i + 1]);
      }
    }
  }
  vector<int> asu = {3, 6, 1, 5, 4, 7, 2};
  for (int i = 0; i < asu.size(); i++) ans[i + 1] = asu[i];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  int n;
  scanf("%d", &n);
  solve_1(n);
  solve_2(n);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int ans[N];
void solve(int n) {
  if (n == 0) return;
  for (int j = 20; j >= 0; j--) {
    if ((1 << j) > n) continue;
    int cen = (1 << j);
    int ada = n - cen + 1;
    for (int i = 0; i < ada; i++) {
      swap(ans[cen + i], ans[cen - i - 1]);
    }
    return solve(cen - ada - 1);
  }
  assert(0);
}
void solve_1(int n) {
  if (n % 2) {
    puts("NO");
    return;
  }
  puts("YES");
  for (int i = 1; i <= n; i++) ans[i] = i;
  solve(n);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
}
void solve_2(int n) {
  if (n <= 5) {
    puts("NO");
    return;
  }
  if (__builtin_popcount(n) == 1) {
    puts("NO");
    return;
  }
  puts("YES");
  if (n == 6) {
    puts("3 6 1 5 4 2");
    return;
  }
  if (n == 7) {
    puts("3 6 1 5 4 7 2");
    return;
  }
  for (int i = 1; i <= n; i++) ans[i] = i;
  for (int j = 17; j >= 3; j--) {
    if ((1 << j) > n) continue;
    int from = (1 << j);
    int to = min(n, from * 2 - 1);
    for (int i = from; i <= to; i++) {
      if (i + 2 <= to) {
        swap(ans[i], ans[i + 2]);
      } else if (i + 1 <= to) {
        swap(ans[i], ans[i + 1]);
      }
    }
  }
  vector<int> asu = {3, 6, 1, 5, 4, 7, 2};
  for (int i = 0; i < asu.size(); i++) ans[i + 1] = asu[i];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  int n;
  scanf("%d", &n);
  solve_1(n);
  solve_2(n);
  return 0;
}

```