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;

int main(){
    int N;
    cin >> N;
    int L = N&(-N);
    if(N == L){
        cout << "No" << endl;
        return 0;
    }

    vector<pair<int, int>> ans;
    int M = N - (N%2 == 0);
    for(int i=2; i<=M; i++){
        ans.emplace_back(1, i);
        ans.emplace_back(N+i, i^1);
    }
    ans.emplace_back(N+1, N+3);
    if(N%2 == 0){
        ans.emplace_back(N, N-L+1);
        ans.emplace_back(2*N, L);
    }

    cout << "Yes" << endl;
    for(auto& p : ans) cout << p.first << " " << p.second << endl;
    return 0;
}