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 <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>

using namespace std;

int main()
{
	int N; scanf("%d",&N);
	int t;
	for(int i = 1;i <= N;t = i,i <<= 1)
		if(N == i)
		{
			printf("No\n");
			return 0;
		}
	if(N == 3)
	{
		printf("Yes\n");
		for(int i = 1;i <= 5; ++ i)
			printf("%d %d\n",i,i + 1);
	}
	else
	{
		puts("Yes");
		for(int i=2; i<N; 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&1))
		{
			printf("%d %d\n", N, N^t);
			printf("%d %d\n", N*2, N+(t^1));
		}
	}
}