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>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  if (n < 3)
    cout << "No" << '\n';
  else {
    cout << "Yes" << '\n';
    vector<int> v1, v2;
    for (int i = (1); i < (n + 1); ++i)
      if (i & 1)
        v1.push_back(i);
      else
        v2.push_back(i);
    cout << v1.size() << " ";
    for (int x : v1) cout << x << " ";
    cout << '\n';
    cout << v2.size() << " ";
    for (int x : v2) cout << x << " ";
    cout << '\n';
  }
  return 0;
}