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<cstdio>
#include<iostream>
#include<vector>
using namespace std;
typedef pair<int,int> PII;
int n;
int pos[400050],rnk[400050];
vector<PII> an;
int main()
{
	scanf("%d",&n);
	int k=0;
	while((1<<k)<n)
	{
		++k;
	}
	if(n==(1<<k))
	{
		printf("No\n");
		return 0;
	}
	pos[1]=1;
	an.push_back(make_pair(n+1,3));
	for(int i=1;i<k;++i)
	{
		int m=1<<i;
		an.push_back(make_pair(1<<(i-1),m));
		an.push_back(make_pair(m+n,m+1+n)); 
		for(int j=1;j<m;++j)
		{
			rnk[pos[j]]=j;
		}
		for(int j=1;j<m&&j+m<=n;++j)
		{
			an.push_back(make_pair(m+j+n,rnk[j]));
			an.push_back(make_pair(m+j,m));
			pos[j+m]=j;
			pos[j]=pos[j]+m; 
		}
		pos[m]=m;
	}
	printf("Yes\n");
	for(unsigned int i=0;i<an.size();++i)
	{
		printf("%d %d\n",an[i].first,an[i].second);
	}
	return 0;
}