Problem: You are given an integer N. Determine if there exists a tree with 2N vertices numbered 1 to 2N satisfying the following condition, and show one such tree if the answer is yes.

* Assume that, for each integer i between 1 and N (inclusive), Vertex i and N+i have the weight i. Then, for each integer i between 1 and N, the bitwise XOR of the weights of the vertices on the path between Vertex i and N+i (including themselves) is i.

Constraints

* N is an integer.
* 1 \leq N \leq 10^{5}

Input

Input is given from Standard Input in the following format:


N


Output

If there exists a tree satisfying the condition in the statement, print `Yes`; otherwise, print `No`. Then, if such a tree exists, print the 2N-1 edges of such a tree in the subsequent 2N-1 lines, in the following format:


a_{1} b_{1}
\vdots
a_{2N-1} b_{2N-1}


Here each pair (a_i, b_i) means that there is an edge connecting Vertex a_i and b_i. The edges may be printed in any order.

Examples

Input

3


Output

Yes
1 2
2 3
3 4
4 5
5 6


Input

1


Output

No

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int N; cin >> N;
    if (__builtin_popcount(N) != 1) {
        cout << "Yes\n";
        for (int i = 2; i+1 <= N; i += 2) {
            cout << i << ' ' << i+1 << '\n';
            cout << i+1 << ' ' << 1 << '\n';
            cout << 1 << ' ' << N+i << '\n';
            cout << N+i << ' ' << N+i+1 << '\n';
            if (i == 2) cout << N+i+1 << ' ' << N+1 << '\n';
        }
        if (N % 2 == 0) {
            cout << N << ' ' << N+N-2 << '\n';
            cout << (N^(N-2)^1) << ' ' << N+N << '\n';
        }
    } else {
        cout << "No\n";
    }
    return 0;
}