#include<cstdio>

using namespace std;

int main(){
	int n;
	scanf("%d",&n);
	int z=1;
	while(z*2<=n)z*=2;
	if(n==z){
		printf("No\n");
		return 0;
	}
	printf("Yes\n");
	for(int i=2;i<n;i+=2){
		printf("%d %d\n",1,i);
		printf("%d %d\n",i,i+1+n);
		printf("%d %d\n",1,i+1);
		printf("%d %d\n",i+1,i+n);
	}
	printf("%d %d\n",n+1,2+n);
	if(n%2==0){
		printf("%d %d\n",z+1,n);
		printf("%d %d\n",n-z,2*n);
	}
}
### Prompt

Your challenge is to write 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<cstdio>

using namespace std;

int main(){
	int n;
	scanf("%d",&n);
	int z=1;
	while(z*2<=n)z*=2;
	if(n==z){
		printf("No\n");
		return 0;
	}
	printf("Yes\n");
	for(int i=2;i<n;i+=2){
		printf("%d %d\n",1,i);
		printf("%d %d\n",i,i+1+n);
		printf("%d %d\n",1,i+1);
		printf("%d %d\n",i+1,i+n);
	}
	printf("%d %d\n",n+1,2+n);
	if(n%2==0){
		printf("%d %d\n",z+1,n);
		printf("%d %d\n",n-z,2*n);
	}
}
```