Problem: 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.

Examples

Input

3


Output

Yes
1 2
2 3
3 4
4 5
5 6


Input

1


Output

No

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

void fail(){
    cout << "No" << endl;
    exit(0);
}

int main(){
    int N;
    cin >> N;

    int N2 = N;
    while(N2 % 2 == 0) N2 /= 2;
    if(N2 == 1) fail();

    vector<pair<int, int>> ans;
    ans.emplace_back(1, 2);
    ans.emplace_back(2, 3);
    ans.emplace_back(3, 1+N);
    ans.emplace_back(1+N, 2+N);
    ans.emplace_back(2+N, 3+N);
    for(int i=4; i<N; i+=2){
        ans.emplace_back(1, i);
        ans.emplace_back(i, i+1);
        ans.emplace_back(1, i+1+N);
        ans.emplace_back(i+1+N, i+N);
    }
    if(N%2 == 0){
        int need = N ^ (N-1);
        need--;
        ans.emplace_back(N, 2*N-1);
        ans.emplace_back(need, 2*N);
    }
    cout << "Yes" << endl;
    for(auto& p : ans) cout << p.first << " " << p.second << endl;
    return 0;
}