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;
template <typename T>
void read(T& x) {
  x = 0;
  char c;
  int sign = 1;
  do {
    c = getchar();
    if (c == '-') sign = -1;
  } while (!isdigit(c));
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (isdigit(c));
  x *= sign;
}
const int N = 1e5 + 500;
int n, m, ans, a[N], f[N];
vector<int> to[N];
int solve(int n) {
  for (register int i = 20; i >= 0; --i)
    if ((n & (1 << i))) {
      int x = ((1 << i + 1) - 1) ^ n;
      if (x > 1) solve(x - 1);
      for (register int j = n; j >= x; --j) printf("%d ", j);
      break;
    }
}
int main() {
  read(n);
  if (n & 1)
    puts("NO");
  else {
    puts("YES");
    solve(n);
    puts("");
  }
  if (n < 6 || n == (n & (-n)))
    puts("NO");
  else {
    puts("YES");
    a[1] = 3;
    a[2] = 6;
    a[3] = 2;
    a[4] = 5;
    a[5] = 1;
    a[6] = 4;
    for (register int i = 7; i <= n; ++i)
      if (i == (i & (-i)))
        a[i] = i + 1, a[i + 1] = i, ++i;
      else
        a[i] = a[(i & (-i))], a[(i & (-i))] = i;
    for (register int i = 1; i <= n; ++i) printf("%d ", a[i]);
  }
  return 0;
}