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<bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long LL;

const int maxn=2e5+5;

int n;

int main()
{
	scanf("%d",&n);
	if (n==1) {puts("No"); return 0;}
	
	int t=0;
	for(t=28; (1<<t)>n; t--);
	if ((1<<t)==n) {puts("No"); return 0;}
	
	puts("Yes");
	for(int i=2; i+1<=n; i+=2)
	{
		printf("%d %d\n",i,i+1);
		printf("%d %d\n",i+1,1);
		printf("%d %d\n",1,n+i);
		printf("%d %d\n",n+i,n+i+1);
	}
	printf("2 %d\n",1+n);
	
	if (!(n&1))
	{
		printf("%d %d\n",n,(1<<t)+1);
		printf("%d %d\n",n-(1<<t)+n,n+n);
	}
}