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 <string>
#include <vector>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
using namespace std;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;

int main()
{
    ll n;
    cin >> n;

    double l = log2(n);
    if (l - floor(l) == 0)
    {
        cout << "No" << endl;
    }
    else
    {
        cout << "Yes" << endl;

        printf("%lld %lld\n", 1, 2);
        printf("%lld %lld\n", 2, 3);
        printf("%lld %lld\n", 3, 1 + n);
        printf("%lld %lld\n", 1 + n, 2 + n);
        printf("%lld %lld\n", 2 + n, 3 + n);

        for (ll i = 4; i < n; i += 2)
        {
            printf("%lld %lld\n", i, i + 1);
            printf("%lld %lld\n", i + 1, 1);
            printf("%lld %lld\n", 1, i + n);
            printf("%lld %lld\n", i + n, i + 1 + n);
        }

        if (n % 2 == 0)
        {
            int x = n + 1;
            for (size_t i = 2; i < n && n <= x; i++)
            {
                x = (n + 1) ^ i;
            }

            printf("%lld %lld\n", x, n);
            printf("%lld %lld\n", (n + 1) ^ x, n + n);
        }
    }

    return 0;
}