Problem: Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:

gcd(sum(S_1), sum(S_2)) > 1 

Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).

Every integer number from 1 to n should be present in exactly one of S_1 or S_2.

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)

Output

If such partition doesn't exist, print "No" (quotes for clarity).

Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.

Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.

If there are multiple possible partitions — print any of them.

Examples

Input

1


Output

No

Input

3


Output

Yes
1 2
2 1 3 

Note

In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".

In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = (int)2e5 + 2;
const int MOD = (int)1e9 + 7;
const bool MULTI_TEST = false;
using namespace std;
void solve() {
  int n;
  cin >> n;
  if (n < 3)
    cout << "No\n";
  else {
    if (n % 2 == 0 || (n % 2 == 1 && (n / 2) % 2 == 1)) {
      cout << "Yes\n";
      cout << n / 2 << ' ';
      for (int i = 2; i <= n; i += 2) cout << i << ' ';
      cout << endl << n - n / 2 << ' ';
      for (int i = 1; i <= n; i += 2) cout << i << ' ';
    } else {
      cout << " Yes\n";
      cout << n - 1 << ' ';
      for (int i = 1; i <= n / 2; i++) cout << i << ' ' << n - i + 1 << ' ';
      cout << endl << 1 << ' ' << n / 2 + 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  t = 1;
  if (MULTI_TEST) cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}