#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

int n;

void init()
{
	scanf("%d",&n);
}

void work()
{
	if(!(n&n-1))
		puts("No");
	else
	{
		puts("Yes");
		puts("1 2");puts("2 3");
		printf("3 %d\n",n+1);
		printf("%d %d\n",n+1,n+2);
		printf("%d %d\n",n+2,n+3);
		for (int i=4;i<n;i+=2)
			printf("1 %d\n1 %d\n%d %d\n%d %d\n",i,i+1,i,n+i+1,i+1,n+i);
		if(!(n&1))
		{
			for (int i=4;i<=n;i++)
			{
				int j=n^1^i;
				if(j<n&&j!=3)
				{
					printf("%d %d\n%d %d\n",i,n,j,n+n);
					break;
				}
			}
		}
	}
}

int main()
{
	init();
	work();
	return 0;
}
### Prompt

Please create a solution in cpp 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<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

int n;

void init()
{
	scanf("%d",&n);
}

void work()
{
	if(!(n&n-1))
		puts("No");
	else
	{
		puts("Yes");
		puts("1 2");puts("2 3");
		printf("3 %d\n",n+1);
		printf("%d %d\n",n+1,n+2);
		printf("%d %d\n",n+2,n+3);
		for (int i=4;i<n;i+=2)
			printf("1 %d\n1 %d\n%d %d\n%d %d\n",i,i+1,i,n+i+1,i+1,n+i);
		if(!(n&1))
		{
			for (int i=4;i<=n;i++)
			{
				int j=n^1^i;
				if(j<n&&j!=3)
				{
					printf("%d %d\n%d %d\n",i,n,j,n+n);
					break;
				}
			}
		}
	}
}

int main()
{
	init();
	work();
	return 0;
}
```