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 <iostream>
#include <algorithm>
#include <vector>
#include <tuple>
using namespace std;

int main(){
    int N;
    cin>>N;
    if(__builtin_popcount(N)==1){
        cout<<"No"<<endl;
        return 0;
    }else{
        cout<<"Yes"<<endl;
    }
    cout<<1<<' '<<2<<endl;
    for(int i=2;i<N;i+=2){
        cout<<i<<' '<<i+1<<endl;
        cout<<i+1<<' '<<1+N<<endl;
        cout<<1+N<<' '<<i+N<<endl;
        cout<<i+N<<' '<<i+1+N<<endl;
    }
    if(N%2==0){
        int y=1;
        while(y<=N)y<<=1;
        y>>=1;
        int x=N^y^1;
        cout<<x<<' '<<N<<endl;
        cout<<y+N<<' '<<2*N<<endl;
    }
    return 0;
}