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 N;

void addEdge(int n, int m){
    cout << n << " " << m << "\n";
}

int main(){
    cin >> N;
    if((1<<__builtin_ctz(N)) == N){
        cout << "No" << "\n";
        return 0;
    }
    cout << "Yes" << "\n";

    for(int i = 2; i<N; i+=2){
        addEdge(i, 1);
        addEdge(i+1, 1);
        addEdge(i, i+1+N);
        addEdge(i+1, i+N);
    }
    addEdge(N+2, N+1);
    if(N&1^1){
        int b = 31 - __builtin_clz(N);
        addEdge(N, (1<<b));
        addEdge(N+N, (1<<b)^N^1);
    }
}