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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define SZ(x) ((int)((x).size()))

template <typename T1, typename T2>
string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
    bool done_something = false;
    stringstream res;
    res << "[";
    for (; begin_iter != end_iter and counter; ++begin_iter) {
        done_something = true;
        counter--;
        res << *begin_iter << ", ";
    }
    string str = res.str();
    if (done_something) {
        str.pop_back();
        str.pop_back();
    }
    str += "]";
    return str;
}

vector<int> SortIndex(int size, std::function<bool(int, int)> compare) {
    vector<int> ord(size);
    for (int i = 0; i < size; i++) ord[i] = i;
    sort(ord.begin(), ord.end(), compare);
    return ord;
}

template <typename T>
bool MinPlace(T& a, const T& b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template <typename T>
bool MaxPlace(T& a, const T& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template <typename S, typename T>
ostream& operator <<(ostream& out, const pair<S, T>& p) {
    out << "{" << p.first << ", " << p.second << "}";
    return out;
}

template <typename T>
ostream& operator <<(ostream& out, const vector<T>& v) {
    out << "[";
    for (int i = 0; i < (int)v.size(); i++) {
        out << v[i];
        if (i != (int)v.size()-1) out << ", ";
    }
    out << "]";
    return out;
}

template<class TH>
void _dbg(const char* name, TH val){
    clog << name << ": " << val << endl;
}
template<class TH, class... TA>
void _dbg(const char* names, TH curr_val, TA... vals) {
    while(*names != ',') clog << *names++;
    clog << ": " << curr_val << ", ";
    _dbg(names+1, vals...);
}

#if DEBUG && !ONLINE_JUDGE
    ifstream input_from_file("input.txt");
    #define cin input_from_file

    #define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
    #define dbg_arr(x, len) clog << #x << ": " << print_iterable(x, x+len, -1) << endl;
#else
    #define dbg(...)
    #define dbg_arr(x, len)
#endif

///////////////////////////////////////////////////////////////////////////
//////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
///////////////////////////////////////////////////////////////////////////


int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); // Remove in problems with online queries!

    int N;
    cin >> N;
    if (__builtin_popcount(N) == 1) {
        cout << "No\n";
        return 0;
    }
    cout << "Yes\n";
    cout << N+1 << " " << 2 << "\n";
    for (int i = 2; i < N; i += 2) {
        cout << i << " " << i + 1 << "\n";
        cout << i+1 << " " << 1 << "\n";
        cout << 1 << " " << i + N << "\n";
        cout << i + N << " " << i + N + 1 << "\n";
    }
    if (N%2 == 0) {
        int pot = 1;
        while ((N&pot) == 0) pot *= 2;
        cout << N << " " << pot + 1 << "\n";
        cout << 2*N << " " << (N^pot)+1 << "\n";
    }
}