#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
const int IT_MAX = 1 << 18;
const int MAXN = 1e5 + 5;
const int mod = 1e9 + 7;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int n, a[MAXN];
void p(int x) {
  if (!x) return;
  int b, i;
  for (b = 1; b <= x; b *= 2)
    ;
  for (i = b - 1 - x; i <= x; i++) a[i] = b - 1 - i;
  p(b - 2 - x);
}
int main() {
  cin >> n;
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    p(n);
    for (int i = 1; i <= n; i++) printf("%d%c", a[i], (i < n) ? ' ' : '\n');
  }
  int b = 1;
  while ((b << 1) <= n) b <<= 1;
  if (n == b || n < 6)
    puts("NO");
  else {
    puts("YES");
    if (n == 6)
      puts("3 6 2 5 1 4");
    else {
      printf("7 3 6 5 1 2 4 ");
      for (int i = 8; i <= n;) {
        int j = min(i << 1, n + 1);
        for (int k = i; k < j - 1; k++) printf("%d ", k + 1);
        printf("%d ", i);
        i = j;
      }
    }
  }
  return 0;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
#pragma warning(disable : 4996)
const int IT_MAX = 1 << 18;
const int MAXN = 1e5 + 5;
const int mod = 1e9 + 7;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int n, a[MAXN];
void p(int x) {
  if (!x) return;
  int b, i;
  for (b = 1; b <= x; b *= 2)
    ;
  for (i = b - 1 - x; i <= x; i++) a[i] = b - 1 - i;
  p(b - 2 - x);
}
int main() {
  cin >> n;
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    p(n);
    for (int i = 1; i <= n; i++) printf("%d%c", a[i], (i < n) ? ' ' : '\n');
  }
  int b = 1;
  while ((b << 1) <= n) b <<= 1;
  if (n == b || n < 6)
    puts("NO");
  else {
    puts("YES");
    if (n == 6)
      puts("3 6 2 5 1 4");
    else {
      printf("7 3 6 5 1 2 4 ");
      for (int i = 8; i <= n;) {
        int j = min(i << 1, n + 1);
        for (int k = i; k < j - 1; k++) printf("%d ", k + 1);
        printf("%d ", i);
        i = j;
      }
    }
  }
  return 0;
  return 0;
}

```