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 PB push_back
#define MP std::make_pair
#define FI first
#define SE second
#define M 1000005
#define INF 1000000000
int n;
int read(){
	char c=getchar();int ans=0;
	while (c<'0'||c>'9') c=getchar();
	while (c>='0'&&c<='9') ans=ans*10+c-'0',c=getchar();
	return ans;
}
void Write(int x){
	if (x<10) putchar(x+'0');
	else Write(x/10),putchar(x%10+'0');
	return;
}
void link(int x,int y){return Write(x),putchar(' '),Write(y),putchar('\n'),void();}
int main(){
	n=read();if (!(n&(n-1))) return !printf("No");printf("Yes\n");
	link(1,2),link(2,3),link(3,n+1),link(n+1,n+2),link(n+2,n+3);
	for (int i=4;i<n;i+=2) link(1,i),link(1,i+1),link(i,n+i+1),link(i+1,n+i);
	if (!(n&1)){
		for (int i=4;i<n;i++){
			int j=i^1^n;
			if (j!=3&&j<n) link(i,n),link(j,n*2),exit(0);
		}
		assert(0);
	}
	return 0;
}