#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 7;
int p[MAXN];
void get(int n) {
  for (int i = 1; i <= n; i++) p[i] = i;
  do {
    int flag = 0;
    for (int i = 1; i <= n; i++)
      if (p[i] == i || (p[i] & i) == 0) {
        flag = 1;
        break;
      }
    if (flag) continue;
    for (int i = 1; i <= n; i++) printf("%d%c", p[i], " \n"[i == n]);
  } while (next_permutation(p + 1, p + 1 + n));
}
void deal1(int n) {
  if (n & 1) {
    puts("NO");
    return;
  }
  puts("YES");
  for (int i = 1; i <= n; i++) p[i] = i;
  int bound = n;
  while (bound) {
    int k = 1;
    while (k <= bound) k *= 2;
    k /= 2;
    int cnt = 0;
    for (int i = 0, j = 1; k + i <= bound; i++, j++)
      swap(p[k + i], p[k - j]), cnt += 2;
    bound -= cnt;
  }
  for (int i = 1; i <= n; i++)
    assert((p[i] & i) == 0), printf("%d%c", p[i], " \n"[i == n]);
}
inline int lowbit(int x) { return x & -x; }
void deal2(int n) {
  if (n <= 5 || lowbit(n) == n) {
    puts("NO");
    return;
  }
  puts("YES");
  for (int i = 1; i <= n; i++) p[i] = i;
  if (n & 1) {
    p[1] = 3;
    p[2] = 6;
    p[3] = 1;
    p[4] = 5;
    p[5] = 4;
    p[6] = 7;
    p[7] = 2;
    for (int i = 8; i <= n; i += 2) swap(p[i], p[i + 1]);
  } else {
    p[1] = 3;
    p[2] = 6;
    p[3] = 2;
    p[4] = 5;
    p[5] = 1;
    p[6] = 4;
    for (int i = 7; i <= n; i += 2) swap(p[i], p[i + 1]);
    for (int i = 7; i <= n; i++)
      if (lowbit(i) == i) {
        swap(p[i], p[i + 1]);
        swap(p[i - 1], p[i + 2]);
      }
  }
  for (int i = 1; i <= n; i++)
    assert((p[i] & i) != 0), printf("%d%c", p[i], " \n"[i == n]);
}
int main() {
  int n;
  scanf("%d", &n);
  deal1(n);
  deal2(n);
  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;
const int MAXN = 1e5 + 7;
int p[MAXN];
void get(int n) {
  for (int i = 1; i <= n; i++) p[i] = i;
  do {
    int flag = 0;
    for (int i = 1; i <= n; i++)
      if (p[i] == i || (p[i] & i) == 0) {
        flag = 1;
        break;
      }
    if (flag) continue;
    for (int i = 1; i <= n; i++) printf("%d%c", p[i], " \n"[i == n]);
  } while (next_permutation(p + 1, p + 1 + n));
}
void deal1(int n) {
  if (n & 1) {
    puts("NO");
    return;
  }
  puts("YES");
  for (int i = 1; i <= n; i++) p[i] = i;
  int bound = n;
  while (bound) {
    int k = 1;
    while (k <= bound) k *= 2;
    k /= 2;
    int cnt = 0;
    for (int i = 0, j = 1; k + i <= bound; i++, j++)
      swap(p[k + i], p[k - j]), cnt += 2;
    bound -= cnt;
  }
  for (int i = 1; i <= n; i++)
    assert((p[i] & i) == 0), printf("%d%c", p[i], " \n"[i == n]);
}
inline int lowbit(int x) { return x & -x; }
void deal2(int n) {
  if (n <= 5 || lowbit(n) == n) {
    puts("NO");
    return;
  }
  puts("YES");
  for (int i = 1; i <= n; i++) p[i] = i;
  if (n & 1) {
    p[1] = 3;
    p[2] = 6;
    p[3] = 1;
    p[4] = 5;
    p[5] = 4;
    p[6] = 7;
    p[7] = 2;
    for (int i = 8; i <= n; i += 2) swap(p[i], p[i + 1]);
  } else {
    p[1] = 3;
    p[2] = 6;
    p[3] = 2;
    p[4] = 5;
    p[5] = 1;
    p[6] = 4;
    for (int i = 7; i <= n; i += 2) swap(p[i], p[i + 1]);
    for (int i = 7; i <= n; i++)
      if (lowbit(i) == i) {
        swap(p[i], p[i + 1]);
        swap(p[i - 1], p[i + 2]);
      }
  }
  for (int i = 1; i <= n; i++)
    assert((p[i] & i) != 0), printf("%d%c", p[i], " \n"[i == n]);
}
int main() {
  int n;
  scanf("%d", &n);
  deal1(n);
  deal2(n);
  return 0;
}

```