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 long long mod1 = 19260817;
const long long mod2 = 19660813;
int n;
int ans[1000005];
void solve1() {
  if (n & 1) {
    puts("NO");
    return;
  }
  puts("YES");
  int s = 1;
  while ((s << 1) <= n) s <<= 1;
  for (int i = n; i >= 1; --i) {
    if (s > i) s >>= 1;
    if (ans[i]) continue;
    int tmp = s - (i - s) - 1;
    ans[i] = tmp;
    ans[tmp] = i;
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  printf("\n");
}
void solve2() {
  int s = 1;
  while ((s << 1) <= n) s <<= 1;
  if (n == s) {
    puts("NO");
    return;
  }
  if (n <= 5) {
    puts("NO");
    return;
  }
  puts("YES");
  if (n == 6) {
    printf("3 6 2 5 1 4\n");
    return;
  }
  if (n == 7) {
    printf("3 6 1 5 4 7 2\n");
    return;
  }
  printf("3 6 1 5 4 7 2");
  for (int i = 8, j; i <= n; i = j) {
    j = min(2 * i, n + 1);
    for (int k = i; k < j - 1; ++k) printf(" %d", k + 1);
    printf(" %d", i);
  }
  printf("\n");
}
int main() {
  scanf("%d", &n);
  solve1();
  solve2();
  return 0;
}