Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3

Sample Output 1

Yes
1 2
2 3
3 4
4 5
5 6

The sample output represents the following graph:

Sample Input 2

1

Sample Output 2

No

There is no tree satisfying the condition.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int main(){
    int N;cin>>N;
    vector<pair<int,int>> a(5);
    for(int i=0;i<2;i++){
        a.at(i) = make_pair(i+1,i+2);
        a.at(i+3) = make_pair(i+N+1,i+N+2);
    }
    a.at(2) = make_pair(3, 1+N);
    if(N<3){
        cout<<"No"<<endl;
    }
    else if(N%2!=0){
        cout<<"Yes"<<endl;
        for(int i=4;i<N;i+=2){
            a.push_back(make_pair(i, i+1));
            a.push_back(make_pair(i+N,i+1+N));
            a.push_back(make_pair(i+1, 1+N));
            a.push_back(make_pair(i+N,1+N));
        }
        for(int i=0;i<a.size();i++){
            cout<<a.at(i).first<<" "<<a.at(i).second<<endl;
        }
    }
    else{
        cout<<"Yes"<<endl;
        a.push_back(make_pair(1+N,2*N));
        a.push_back(make_pair(N-1,N));
        for(int i=4;i<=N-1;i+=2){
            a.push_back(make_pair(i, i+1));
            a.push_back(make_pair(i+N,i+1+N));
            a.push_back(make_pair(i, 1+N));
            a.push_back(make_pair(i+N+1,1+N));
        }
        for(int i=0;i<a.size();i++){
        cout<<a.at(i).first<<" "<<a.at(i).second<<endl;
        }
    }
}