#include <bits/stdc++.h>
using namespace std;
int n, s = 1;
int p[100000 + 1], q[100000 + 1];
void solvep(int n, int s) {
  if (n <= 0) return;
  if (s > n) return solvep(n, s >> 1);
  for (int i = 1; s + i - 1 <= n; i++)
    p[s - i] = s + i - 1, p[s + i - 1] = s - i;
  solvep((s << 1) - n - 2, s >> 1);
}
void solveq(int s) {
  if (s > n) return;
  int m = min((s << 1) - 1, n);
  q[s] = m;
  for (int i = s + 1; i <= m; i++) q[i] = i - 1;
  solveq(s << 1);
}
int main() {
  scanf("%d", &n);
  while ((s << 1) <= n) s <<= 1;
  if (n & 1)
    printf("NO\n");
  else {
    solvep(n, s);
    printf("YES\n");
    for (int i = 1; i <= n; i++) printf("%d ", p[i]);
    printf("\n");
  }
  if (n < 6 || n == s)
    printf("NO\n");
  else if (n == 6)
    printf("YES\n3 6 2 5 1 4\n");
  else if (n == 7)
    printf("YES\n7 6 2 5 4 3 1\n");
  else {
    q[1] = 7;
    q[2] = 6;
    q[3] = 2;
    q[4] = 5;
    q[5] = 4;
    q[6] = 3;
    q[7] = 1;
    printf("YES\n");
    solveq(8);
    for (int i = 1; i <= n; i++) printf("%d ", q[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Your task is to create 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 n, s = 1;
int p[100000 + 1], q[100000 + 1];
void solvep(int n, int s) {
  if (n <= 0) return;
  if (s > n) return solvep(n, s >> 1);
  for (int i = 1; s + i - 1 <= n; i++)
    p[s - i] = s + i - 1, p[s + i - 1] = s - i;
  solvep((s << 1) - n - 2, s >> 1);
}
void solveq(int s) {
  if (s > n) return;
  int m = min((s << 1) - 1, n);
  q[s] = m;
  for (int i = s + 1; i <= m; i++) q[i] = i - 1;
  solveq(s << 1);
}
int main() {
  scanf("%d", &n);
  while ((s << 1) <= n) s <<= 1;
  if (n & 1)
    printf("NO\n");
  else {
    solvep(n, s);
    printf("YES\n");
    for (int i = 1; i <= n; i++) printf("%d ", p[i]);
    printf("\n");
  }
  if (n < 6 || n == s)
    printf("NO\n");
  else if (n == 6)
    printf("YES\n3 6 2 5 1 4\n");
  else if (n == 7)
    printf("YES\n7 6 2 5 4 3 1\n");
  else {
    q[1] = 7;
    q[2] = 6;
    q[3] = 2;
    q[4] = 5;
    q[5] = 4;
    q[6] = 3;
    q[7] = 1;
    printf("YES\n");
    solveq(8);
    for (int i = 1; i <= n; i++) printf("%d ", q[i]);
    printf("\n");
  }
  return 0;
}

```