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<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;



int main(){
    int n;
    cin >> n;
    if((n & -n) == n){
        cout << "No" << endl;
        return 0;
    }
    cout << "Yes" << endl;
    int k = n;
    while(k % 4 != 3)k--;
    for(int i = 1;i < k;i++){
        cout << i << " " << i + 1 << endl;
        cout << i + n << " " << i + n + 1 << endl;
    }
    cout << k << " "<< 1 + n << endl;
    if(n % 4 == 3)return 0;
    if(n % 4 == 0){
        int r = 1;
        while(r * 2 <= n)r *= 2;
        int l = n - r;
        cout << n << " " << l + 3 <<endl; 
        cout << n + n << " " << r + 3 <<endl;
        return 0;
    }
    k++;
    cout << k+1 << " " << k << endl;
    cout << k << " " << 1 << endl;
    cout << 1 << " " << k+1+n << endl;
    cout << k+1+n << " " << k+n << endl;
    if(n % 4 == 1)return 0;
    cout << n << " " << 2 << endl;
    cout << n + n << " " << k+1+n << endl;
    return 0;
}