#include <bits/stdc++.h>
using namespace std;
int n, a[200005];
void recur(int base, int bit) {
  if (bit == 0) return;
  swap(a[base + (1 << bit)], a[base + (1 << bit) - 1]);
  for (int i = 0; i < (1 << bit) - 2; i++) {
    if (base + (1 << bit) + i + 1 > n) break;
    swap(a[base + i + 1], a[base + (1 << bit) + i + 1]);
  }
  recur(base, bit - 1);
  recur(base + (1 << bit), bit - 1);
}
void makep() {
  if (n % 2 == 1) {
    printf("NO\n");
    return;
  }
  int bit;
  for (bit = 0; (1 << (bit + 1)) <= n; bit++) {
  }
  for (int i = 1; i <= n; i++) {
    a[i] = i;
  }
  recur(0, bit);
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    if (i > 1) printf(" ");
    printf("%d", a[i]);
  }
  printf("\n");
}
void makeq() {
  if (n <= 5) {
    printf("NO\n");
    return;
  }
  int bit;
  for (bit = 0; (1 << (bit + 1)) <= n; bit++) {
  }
  if (n == (1 << bit)) {
    printf("NO\n");
    return;
  }
  printf("YES\n");
  if (n == 6) {
    printf("3 6 2 5 1 4\n");
    return;
  }
  printf("7 6 2 5 4 3 1");
  for (int b = 3; b <= bit; b++) {
    for (int i = 0; i < (1 << b); i++) {
      if (i == (1 << b) - 1 || (1 << b) + i == n) {
        printf(" %d", (1 << b));
        break;
      } else
        printf(" %d", (1 << b) + i + 1);
    }
  }
  printf("\n");
}
int main() {
  scanf("%d", &n);
  makep();
  makeq();
  return 0;
}

### Prompt

Your challenge is to write 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, a[200005];
void recur(int base, int bit) {
  if (bit == 0) return;
  swap(a[base + (1 << bit)], a[base + (1 << bit) - 1]);
  for (int i = 0; i < (1 << bit) - 2; i++) {
    if (base + (1 << bit) + i + 1 > n) break;
    swap(a[base + i + 1], a[base + (1 << bit) + i + 1]);
  }
  recur(base, bit - 1);
  recur(base + (1 << bit), bit - 1);
}
void makep() {
  if (n % 2 == 1) {
    printf("NO\n");
    return;
  }
  int bit;
  for (bit = 0; (1 << (bit + 1)) <= n; bit++) {
  }
  for (int i = 1; i <= n; i++) {
    a[i] = i;
  }
  recur(0, bit);
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    if (i > 1) printf(" ");
    printf("%d", a[i]);
  }
  printf("\n");
}
void makeq() {
  if (n <= 5) {
    printf("NO\n");
    return;
  }
  int bit;
  for (bit = 0; (1 << (bit + 1)) <= n; bit++) {
  }
  if (n == (1 << bit)) {
    printf("NO\n");
    return;
  }
  printf("YES\n");
  if (n == 6) {
    printf("3 6 2 5 1 4\n");
    return;
  }
  printf("7 6 2 5 4 3 1");
  for (int b = 3; b <= bit; b++) {
    for (int i = 0; i < (1 << b); i++) {
      if (i == (1 << b) - 1 || (1 << b) + i == n) {
        printf(" %d", (1 << b));
        break;
      } else
        printf(" %d", (1 << b) + i + 1);
    }
  }
  printf("\n");
}
int main() {
  scanf("%d", &n);
  makep();
  makeq();
  return 0;
}

```