#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	scanf("%d",&n);
	if(n-(n&-n)==0){printf("No");return 0;}
	printf("Yes\n");
		
	if(n==3)
	{
		printf("1 2\n2 3\n3 4\n4 5\n5 6\n");
		return 0;
	}
	printf("1 2\n2 3\n3 %d\n%d %d\n%d %d\n",n+1,n+1,n+2,n+2,n+3);
	for(int i=4;i+1<=n;i+=2)
	{
		printf("%d %d\n%d %d\n",n+1,i,i,i+n+1);
		printf("%d %d\n%d %d\n",n+1,i+1,i+1,i+n);
	}
	if(n%2==0)
	{
		int k=n&-n;
		int x=n^k^1;
		if(k!=2)
			printf("%d %d\n%d %d\n",k,n,x,n+n);
		else
			printf("%d %d\n%d %d\n",n+2,n,x,n+n);
	}
	return 0;
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	scanf("%d",&n);
	if(n-(n&-n)==0){printf("No");return 0;}
	printf("Yes\n");
		
	if(n==3)
	{
		printf("1 2\n2 3\n3 4\n4 5\n5 6\n");
		return 0;
	}
	printf("1 2\n2 3\n3 %d\n%d %d\n%d %d\n",n+1,n+1,n+2,n+2,n+3);
	for(int i=4;i+1<=n;i+=2)
	{
		printf("%d %d\n%d %d\n",n+1,i,i,i+n+1);
		printf("%d %d\n%d %d\n",n+1,i+1,i+1,i+n);
	}
	if(n%2==0)
	{
		int k=n&-n;
		int x=n^k^1;
		if(k!=2)
			printf("%d %d\n%d %d\n",k,n,x,n+n);
		else
			printf("%d %d\n%d %d\n",n+2,n,x,n+n);
	}
	return 0;
}

```