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;
	if(1 << (int)(log2(N + 0.1)) == N){puts("No"); return 0;}
	puts("Yes");
	for(int i = 1 ; i < 3 ; ++i){
		cout << i << ' ' << i + 1 << endl;
		cout << i + N << ' ' << i + N + 1 << endl;
	}
	cout << 3 << ' ' << N + 1 << endl;
	for(int i = 4 ; i < N ; i += 2){
		cout << i << ' ' << i + 1 << endl;
		cout << i + 1 << ' ' << N + 1 << endl;
		cout << N + 1 << ' ' << i + N << endl;
		cout << i + N << ' ' << i + N + 1 << endl;
	}
	if(!(N & 1)){
		int t = 1 << (int)(log2(N + 0.1));
		cout << N << ' ' << t + N << endl; cout << N - t + 1 << ' ' << 2 * N << endl;
	}
	return 0;
}