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 <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <functional>
#include <string>
#include <cmath>
#include <cassert>

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

int main()
{
	int n;
	scanf("%d",&n);
	int T=n;
	while(T%2==0) T/=2;
	if(T==1) puts("No");
	else
	{
		puts("Yes");
		int z=n;
		if(z%2==0) z--;
		for(int i=2;i<=z;i+=2)
		{
			printf("%d %d\n",1,i);
			printf("%d %d\n",i,i+1);
			printf("%d %d\n",1,n+i+1);
			printf("%d %d\n",n+i+1,n+i);
		}
		printf("%d %d\n",n+1,3);
		if(n%2==0)
		{
			for(int i=20;i>=0;i--)
			{
				if(n>>i&1)
				{
					int A=(1<<i),B=n-A;
					printf("%d %d\n",A,n);
					printf("%d %d\n",B+n+1,2*n);
					break;
				}
			}
		}
	}
	return 0;
}