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>
#include <random>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;
//#define ll __int128
//#define int ll
//#define int ll
//#define char ll
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<vvc> vvvc;
typedef pair<int, int> pii;
typedef pair<pii, pii> piii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;

#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cerr << (#x) << " = " << (x) << endl;

#ifdef MAX_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif

const double PI = 2 * acos(0.0);
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";


template <class T0, class T1>
inline ostream & operator << (ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template <class T0, class T1>
inline istream & operator >> (istream &in, pair<T0, T1> &a) {
    return in >> a.first >> a.second;
}

template <class T0, class T1, class T2>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template <class T0, class T1, class T2, class T3>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << ", " <<  get<3>(a) << "}";
}

template<class T>
inline ostream & operator << (ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())
        out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}


void smain();



signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

#ifdef MAX_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(4) << fixed;
    smain();
#ifdef MAX_HOME
    cout << "\n\n\n\nTOTAL EXECUTION TIME: " << float( clock () - start ) /  CLOCKS_PER_SEC << endl;
#endif
    return 0;
}

const int L = 17;

bool cool[1 << L];

void smain() {

    int xr = 0;
    for (int i = 0; i < (1 << L); ++i) {
        xr ^= i;
        cool[xr] = true;
    }
    fori (mask, (1 << L)) {
        for (int m2 = mask; ; m2 = (m2 - 1) & mask) {
            cool[mask] |= cool[m2] && cool[mask ^ m2];
            if (!m2)
                break;
        }
    }

    int n;
    cin >> n;
    if (n == 1) {
        cout << "No\n" << endl;
        return;
    }
    int big = 0;
    for (int mask = 0; mask <= n; ++mask) {
        if (cool[mask])
            big = mask;
    }
    vector<pii> ans;
    for (int i = 1; i < big; ++i) {
        ans.push_back({i, i + 1});
        ans.push_back({i + n, i + n + 1});
    };
    ans.push_back({big, n + 1});
    for (int i = big + 1; i <= n; ++i) {
        ans.push_back({i - big, i});
        int kek = i - big + n - 1;
        if (i == big + 1)
            kek = big + n;
        ans.push_back({kek, i + n});
    }
    cout << "Yes\n";
    for (auto x : ans) {
        cout << x.first << ' ' << x.second << '\n';
    }
}