#include <bits/stdc++.h>
using namespace std;
int a[100005], b[100005] = {0, 3, 7, 1, 5, 6, 4, 2};
inline int cnt(int x) {
  int rtn = 0;
  while (x) rtn += x & 1, x >>= 1;
  return rtn;
}
inline int pd(int x) {
  if (x < 6) return 0;
  if (cnt(x) == 1) return 0;
  return 1;
}
int main(int argc, char** argv) {
  int n;
  cin >> n;
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    int now = (1 << 30) - 1;
    for (int i = n; i >= 1; i--) {
      if (a[i]) continue;
      while (now - i > i) now >>= 1;
      a[i] = now - i, a[now - i] = i;
    }
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    puts("");
  }
  if (!pd(n))
    puts("NO");
  else {
    puts("YES");
    if (n == 6) {
      cout << "3 6 2 5 1 4";
      return 0;
    }
    if (n == 7) {
      cout << "3 7 1 5 6 4 2";
      return 0;
    }
    int now = 1;
    while ((now << 1) < n) now <<= 1;
    int N = n;
    while (n > 7) {
      for (int i = now; i <= n; i++) b[i] = i + 1;
      b[n] = now;
      n = now - 1;
      now >>= 1;
    }
    for (int i = 1; i <= N; i++) cout << b[i] << " ";
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a[100005], b[100005] = {0, 3, 7, 1, 5, 6, 4, 2};
inline int cnt(int x) {
  int rtn = 0;
  while (x) rtn += x & 1, x >>= 1;
  return rtn;
}
inline int pd(int x) {
  if (x < 6) return 0;
  if (cnt(x) == 1) return 0;
  return 1;
}
int main(int argc, char** argv) {
  int n;
  cin >> n;
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    int now = (1 << 30) - 1;
    for (int i = n; i >= 1; i--) {
      if (a[i]) continue;
      while (now - i > i) now >>= 1;
      a[i] = now - i, a[now - i] = i;
    }
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    puts("");
  }
  if (!pd(n))
    puts("NO");
  else {
    puts("YES");
    if (n == 6) {
      cout << "3 6 2 5 1 4";
      return 0;
    }
    if (n == 7) {
      cout << "3 7 1 5 6 4 2";
      return 0;
    }
    int now = 1;
    while ((now << 1) < n) now <<= 1;
    int N = n;
    while (n > 7) {
      for (int i = now; i <= n; i++) b[i] = i + 1;
      b[n] = now;
      n = now - 1;
      now >>= 1;
    }
    for (int i = 1; i <= N; i++) cout << b[i] << " ";
  }
  return 0;
}

```