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;
#define finish(x) return cout << x << endl, 0
#define ll long long

int n;

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    if((n & (n - 1)) == 0) finish("No");
    cout << "Yes\n";
    cout << "1 2\n2 3\n3 " << n + 1 << "\n" << n + 1 << " " << n + 2 << "\n" << n + 2 << " " << n + 3 << "\n";
    for(int i = 4 ; i + 1 <= n ; i += 2){
        cout << i + 1 << " " << i << "\n" << i << " 1\n1 " << n + i + 1 << "\n" << n + i + 1 << " " << n + i << "\n";
    }
    if(n % 2 == 0){
        for(int i = 2 ; i < n ; i++){
            int j = n ^ 1 ^ i;
            if(1 < j && j < n){
                cout << n << " " << i + n * (i % 2) << "\n" << j + n * (j % 2) << " " << n + n << "\n";
                break;
            }
        }
    }
}