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 llong long long
#define mkpr make_pair
#define riterator reverse_iterator
using namespace std;

inline int read()
{
	int x=0,f=1; char ch=getchar();
	for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}
	for(; isdigit(ch);ch=getchar()) {x = x*10+ch-'0';}
	return x*f;
}

int n;

int main()
{
	scanf("%d",&n); int p = n&(-n);
	if(p==n) {puts("No"); return 0;}
	puts("Yes");
	for(int i=3; i<=n; i+=2)
	{
		printf("%d %d\n%d %d\n%d %d\n%d %d\n",1,i,i,i-1,1,i-1+n,i-1+n,i+n);
	}
	printf("%d %d\n",2,n+1);
	if(!(n&1))
	{
		int x = p>>1,y = (n^p)>>1;
		printf("%d %d\n%d %d\n",n,2*x+1,n+n,2*y+n);
	}
	return 0;
}