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>
#include <algorithm>
using namespace std;
int main(){
	int N; cin >> N; 
	int A[18] = {1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072}; 
	bool possible = true; 
	for(int i = 0; i <= 17; i++){
		if(N == A[i]){
			cout << "No" << endl; return 0; 
		}
	}
	cout << "Yes" << endl; 
	cout << N + 1 << " " << N + 2 << endl; 
	for(int i = 1; i < (N+1) / 2; i++){
		cout << 1 << " " << 2*i << endl; 
		cout << 2*i << " " << 2*i+1 << endl; 
		cout << 1 << " " << 2*i+1+N << endl; 
		cout << 2*i+N << " " << 2*i+1+N << endl;
	}
	if(N % 2 == 0){
		int M = (lower_bound(A, A+18, N) - A) - 1; 
		int a = A[M]; int b = N - A[M]; 
		cout << N << " " << a << endl; 
		cout << N*2 << " " << b + 1 + N << endl; 
	}
}