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 <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <queue>
#include <random>
#include <regex>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <sys/timeb.h>
#include <vector>

using namespace std;

#define repr(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) repr(i, 0, n)
#define reprrev(i, a, b) for (int i = (int)(b)-1; i >= (int)(a); i--)
#define reprev(i, n) reprrev(i, 0, n)
#define repi(itr, ds) for (auto itr = ds.begin(); itr != ds.end(); itr++)
#define chmin(mi, val) mi = min(mi, val)
#define chmax(ma, val) ma = max(ma, val)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define mp make_pair
#define mt make_tuple
#define INF 1050000000
#define INFR INT_MAX
#define INFL (long long)(4e18)
#define INFLR LLONG_MAX
#define EPS (1e-10)
#define MOD 1000000007
// #define MOD 998244353
#define PI 3.141592653589793238
#define RMAX 4294967295

using vi = vector<int>;
using vvi = vector<vector<int>>;
using vvvi = vector<vector<vector<int>>>;
using vvvvi = vector<vector<vector<vector<int>>>>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using vvvll = vector<vector<vector<ll>>>;
using vd = vector<double>;
using vvd = vector<vector<double>>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
using vc = vector<char>;
using vvc = vector<vector<char>>;
using vs = vector<string>;
using vvs = vector<vector<string>>;
using Pi = pair<int, int>;
using vPi = vector<Pi>;
using vvPi = vector<vector<Pi>>;
using vvvPi = vector<vector<vector<Pi>>>;
using vvvvPi = vector<vector<vector<vector<Pi>>>>;
using Pll = pair<ll, ll>;
using vPll = vector<Pll>;
using Pd = pair<double, double>;
using vPd = vector<Pd>;
template <class T>
using vec = vector<T>;
template <class T>
using pql = priority_queue<T, vector<T>, greater<T>>;

// vvvvvvvvvvvvvvvvvvvvvvv debug output vvvvvvvvvvvvvvvvvvvvvvv
// vector
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
    for (T &x : vec) is >> x;
    return is;
}
// pair
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}
// vector
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
    os << "{";
    for (int i = 0; i < vec.size(); i++) {
        os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
    }
    os << "}";
    return os;
}
// map
template <typename T, typename U>
ostream &operator<<(ostream &os, map<T, U> &map_var) {
    os << "{";
    repi(itr, map_var) {
        os << *itr;
        itr++;
        if (itr != map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// set
template <typename T>
ostream &operator<<(ostream &os, set<T> &set_var) {
    os << "{";
    repi(itr, set_var) {
        os << *itr;
        itr++;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}

#define DUMPOUT cerr

void dump_func() {
    DUMPOUT << endl;
}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) > 0) {
        DUMPOUT << ", ";
    }
    dump_func(std::move(tail)...);
}
#ifdef DEBUG_
#define DEB
#define dump(...)                                                              \
    DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                            \
            << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]"        \
            << endl                                                            \
            << "    ",                                                         \
        dump_func(__VA_ARGS__)
#else
#define DEB if (false)
#define dump(...)
#endif

// ^^^^^^^^^^^^^^^^^^^^^^^ debug output ^^^^^^^^^^^^^^^^^^^^^^^

string YN(bool y, int id = 0) {
    if (id) cout << id;
    return (y ? "YES" : "NO");
}
string yn(bool y, int id = 0) {
    if (id) cout << id;
    return (y ? "Yes" : "No");
}
string ON(bool y, int id = 0) {
    if (id) cout << id;
    return (y ? "OK" : "NG");
}

int dir4[4][2] = {{0, -1}, {-1, 0}, {1, 0}, {0, 1}};
int dir8[8][2] = {{-1, -1}, {0, -1}, {1, -1}, {-1, 0},
                  {1, 0},   {-1, 1}, {0, 1},  {1, 1}};
char dirchar[4] = {'<', '^', '>', 'v'};

// [a,b)
int irand(int a, int b) {
    static mt19937 Rand(static_cast<unsigned int>(time(nullptr)));
    uniform_int_distribution<int> dist(a, b - 1);
    return dist(Rand);
}

// [a,b)
double drand(int a, int b) {
    static mt19937 Rand(static_cast<unsigned int>(time(nullptr)));
    uniform_real_distribution<double> dist(a, b);
    return dist(Rand);
}

// https://qiita.com/IgnorantCoder/items/3101d6276e9bdddf872c
template <typename A, typename F>
inline auto transform(const A &v, F &&f) {
    using result_type =
        decltype(std::declval<F>()(std::declval<typename A::value_type>()));
    vector<result_type> y(v.size());
    std::transform(std::cbegin(v), std::cend(v), std::begin(y), f);
    return y;
}

// 多次元vector生成
template <class T>
vector<T> make_v(size_t size, const T &init) {
    return vector<T>(size, init);
}
template <class... Ts>
auto make_v(size_t size, Ts... rest) {
    return vector<decltype(make_v(rest...))>(size, make_v(rest...));
}

template <typename T>
T Max(vector<T> a) {
    return *max_element(all(a));
}
template <typename T>
T Min(vector<T> a) {
    return *min_element(all(a));
}
template <typename T>
T Sum(vector<T> a) {
    return accumulate(all(a), (T)0);
}

// mapでカウントとかする
template <typename T>
void Add(map<T, int> &m, T item) {
    if (m.find(item) == m.end()) {
        m[item] = 1;
    } else {
        m[item]++;
    }
}

// デフォルト値つきのmapのget
template <typename T, typename U>
U Get(map<T, U> m, T key, U def) {
    if (m.find(key) == m.end()) {
        return def;
    } else {
        return m[key];
    }
}

template <typename T>
bool Contains(set<T> t, const T &key) {
    return t.find(key) != t.end();
}

template <typename T, typename U>
bool Contains(map<T, U> t, const T &key) {
    return t.find(key) != t.end();
}

template <class T>
struct Edge {
    int from, to;
    T cost;
    bool operator<(Edge e) {
        return cost < e.cost;
    }
};

template <class T>
ostream &operator<<(ostream &os, Edge<T> &edge) {
    os << "(" << edge.from << "->" << edge.to << ":" << edge.cost << ")";
    return os;
}
template <class T = int>
class Graph {
    int n;
    bool directed;
    vector<vector<Edge<T>>> edges;

  public:
    Graph(int n, bool directed = false)
        : n(n), directed(directed), edges(vector<vector<Edge<T>>>(n)) {}

    void add_edge(int s, int t, T cost) {
        edges[s].push_back(Edge<T>{s, t, cost});
        if (!directed) {
            edges[t].push_back(Edge<T>{t, s, cost});
        }
    }
    Graph() {}

    vector<Edge<T>> operator[](size_t i) const {
        return edges[i];
    }

    int size() const {
        return n;
    }
};

//======================================================

int digit(int n) {
    int ans = 0;
    while (n > 0) {
        n >>= 1;
        ans++;
    }
    return ans;
}

void answer(const vPi &edges) {
    cout << "Yes" << endl;
    dump(edges.size());
    for (auto p : edges) {
        cout << p.first << " " << p.second << endl;
    }
    return;
}

int main(void) {
    int N;
    cin >> N;
    if (N <= 2) {
        cout << "No" << endl;
        return 0;
    }
    vPi edges;
    edges.reserve(N - 1);

    edges.push_back(mp(1, 2));
    edges.push_back(mp(2, 3));
    edges.push_back(mp(3, N + 1));
    edges.push_back(mp(N + 1, N + 2));
    edges.push_back(mp(N + 2, N + 3));

    if (N == 3) {
        answer(edges);
        return 0;
    }

    repr(i, 1, (N + 1) / 4) {
        edges.push_back(mp(i * 4, i * 4 + 1));
        edges.push_back(mp(i * 4 + 1, i * 4 + 2));
        edges.push_back(mp(i * 4 + 2, i * 4 + 3));
        edges.push_back(mp(i * 4 + 3, N + i * 4));
        edges.push_back(mp(N + i * 4, N + i * 4 + 1));
        edges.push_back(mp(N + i * 4 + 1, N + i * 4 + 2));
        edges.push_back(mp(N + i * 4 + 2, N + i * 4 + 3));
    }

    if (N % 4 == 3) {
        // 並べるだけ
        repr(i, 1, (N + 3) / 4) {
            // i*4 - i*4+3 のブロック
            edges.push_back(mp(1, i * 4));
        }
    } else if (N % 4 == 0) {
        // 2数で条件を満たす数を探す
        // 無理な場合がある
        if (digit(N) != digit(N - 1)) {
            cout << "No" << endl;
            return 0;
        }
        int target = N ^ (N - 1);
        edges.push_back(mp(target, N - 1));
        edges.push_back(mp(N, target));
        edges.push_back(mp(N - 1, 2 * N));
        repr(i, 1, N / 4 - 1) {
            // i*4 - i*4+3 のブロック
            edges.push_back(mp(1, i * 4));
        }
    } else if (N % 4 == 1) {
        // 間の数を計算して差し込む
        int target = N ^ (N - 1);
        edges.push_back(mp(N - 1, N));
        edges.push_back(mp(N, target));
        edges.push_back(mp(target, N + N - 1));
        edges.push_back(mp(N + N - 1, N + N));
        repr(i, 1, N / 4) {
            // i*4 - i*4+3 のブロック
            edges.push_back(mp(1, i * 4));
        }
    } else if (N % 4 == 2) {
        // 無理な場合があるっぽい
        if (N == 6) {
            edges.push_back(mp(4, 5));
            edges.push_back(mp(5, 1));
            edges.push_back(mp(1, 10));
            edges.push_back(mp(10, 11));
            edges.push_back(mp(6, 5));
            edges.push_back(mp(1, 12));
            answer(edges);
            return 0;
        }
        if (digit(N) != digit(N - 3)) {
            edges.push_back(mp(N - 2, N));
            edges.push_back(mp(N, 2));
            edges.push_back(mp(2, N + N - 2));
            edges.push_back(mp(N + N - 2, N + N));

            edges.push_back(mp(N - 1, N - 2));
            edges.push_back(mp(N - 2, N - 4));
            edges.push_back(mp(N - 3, N + N - 1));
            repr(i, 1, N / 4 - 1) {
                // i*4 - i*4+3 のブロック
                edges.push_back(mp(1, i * 4));
            }
        } else {
            int target = N ^ (N - 1) ^ (N - 2) ^ (N - 3);
            edges.push_back(mp(N - 2, N - 1));
            edges.push_back(mp(N - 1, N));
            edges.push_back(mp(N, target));
            edges.push_back(mp(target, N + N - 3));
            edges.push_back(mp(N + N - 3, N + N - 2));
            edges.push_back(mp(N + N - 2, N + N - 1));
            edges.push_back(mp(N + N - 1, N + N));
            repr(i, 1, N / 4 - 1) {
                // i*4 - i*4+3 のブロック
                edges.push_back(mp(1, i * 4));
            }
        }
    }
    answer(edges);

    return 0;
}