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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

int main(){
	int n;cin>>n;
	if(__builtin_popcount(n)==1){
		puts("No");return 0;
	}
	puts("Yes");
	vector<P>v;
	int a[]{1+n,2,3,1,2+n,3+n};
	rep(i,5)v.push_back(P(a[i],a[i+1]));
	for(int i=4;i+1<=n;i+=2){
		v.push_back(P(i,i+1));
		v.push_back(P(i+1,1));
		v.push_back(P(1,n+i));
		v.push_back(P(n+i,n+i+1));
	}
	if(n%2==0){
		int x=n^(n-1)^1;
		v.push_back(P(n,n-1));
		if(x%2==0){
			v.push_back(P(x+n,n*2));
		}
		else{
			v.push_back(P(x,n*2));
		}
	}
	for(auto p:v){
		printf("%d %d\n",p.first,p.second);
	}
}