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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int ddx[] = {-1, -2, 1, -2, 2, -1, 2, 1}, ddy[] = {-2, -1, -2, 1, -1, 2, 1, 2};
bool chk[100005];
int ans[100005];
int main() {
  int n;
  for (int(i) = (0); (i) <= (16); (i) += (1)) chk[1 << i] = 1;
  scanf("%d", &n);
  if (n & 1) {
    puts("NO");
  } else {
    puts("YES");
    priority_queue<int> q;
    for (int(i) = (1); (i) <= (n); (i) += (1)) q.push(i);
    for (int(i) = (1); (i) <= (n / 2); (i) += (1)) {
      while (!q.empty() && ans[q.top()]) q.pop();
      int x = q.top();
      q.pop();
      int t = 32 - __builtin_clz(x);
      int y = ((1 << t) - 1) ^ x;
      ans[x] = y;
      ans[y] = x;
    }
    for (int(i) = (1); (i) <= (n); (i) += (1)) printf("%d ", ans[i]);
    puts("");
  }
  if (n <= 5 || chk[n]) {
    puts("NO");
  } else {
    puts("YES");
    if (n == 6) {
      puts("3 6 2 5 1 4");
    } else {
      printf("3 6 1 5 4 7 2 ");
      for (int(i) = (8); (i) <= (n); (i) += (1)) {
        if (chk[i])
          printf("%d ", min(2 * i - 1, n));
        else
          printf("%d ", i - 1);
      }
    }
  }
}