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<iostream>
#include<cstdio>
using namespace std;
int n;
int main()
{
    // freopen("toad.in","r",stdin);
    // freopen("toad.out","w",stdout);
	scanf("%d",&n);
	int m=1;
	while(m<n) m<<=1;
	if(n<3||m==n)
    {
        printf("No");
        return 0;
    }
	printf("Yes\n");
    printf("%d %d\n",1,n+2);
    printf("%d %d\n",n+2,3);
    printf("%d %d\n",3,n+1);
    printf("%d %d\n",n+1,2);
    printf("%d %d\n",2,n+3);
    for(int i=4;i<n;i+=2)
    {
        printf("%d %d\n",i,n+1);
        printf("%d %d\n",i,n+i+1);
        printf("%d %d\n",i+1,n+1);
        printf("%d %d\n",n+i,i+1);
    }
    if(n%2==0)
    {
        for(int i=2;i<n;i++)
            if((n^(i^1))<n)
            {
                printf("%d %d\n",n,i);
                printf("%d %d\n",n+n,n^i^1);
                break;
            }
	}
	return 0;
}