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;

vector<pair<int,int> >v;

int main()
{
    int n;
    scanf("%d",&n);
    if(n<3 || __builtin_popcount(n)==1)
        puts("No");
    else
    {
        puts("Yes");
        v.push_back({1,2});
        v.push_back({2,3});
        v.push_back({3,n+1});
        v.push_back({n+1,n+2});
        v.push_back({n+2,n+3});

        for(int i=4;i<n;i+=2)
        v.push_back({n+1,i}),v.push_back({i,i+1});

        for(int i=n+5;i<=2*n;i+=2)
        v.push_back({n+1,i}),v.push_back({i,i-1});


        if(n%2==0)
        {
            int x,y;
            for(int i=0;;i++)
            {
                if(1<<i&n)
                {
                    x=1<<i;break;
                }
            }
            y=n^x^1;
           v.push_back({x,n});
           v.push_back({n+y,2*n});
        }
        for(auto i :v)
            printf("%d %d\n",i.first,i.second);

    }

    return 0;
}