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> 
using namespace std;
 
 
int main()
{ int n,i,sum=0,m,s=0,x; scanf("%d",&n);
if(n==1){printf("No"); return 0;}
m=n;
while(m%2==0){m/=2;}
if(m==1){printf("No"); return 0;}
m=n+1;
while(m){m/=2; s++;}  s--;
m=1; for(i=0;i<s;i++){m*=2;} int A[m];
for(i=0;i<m;i++){A[i]=i;}


printf("Yes\n");
//printf("%d\n",m);
for(i=m;i<n;i+=2){
printf("%d %d\n",i,i+1);
printf("1 %d\n",i);
printf("1 %d\n",i+n+1);
printf("%d %d\n",i+n,i+1+n);	
	
} 
if(n%2==0){
printf("%d %d\n",n,n-2);	
x=n^(n-2); x=x^1;
A[x]=2; A[2]=x;
printf("%d %d\n",2*n,x);	

	
	
	
}
for(i=1;i<m-1;i++){
printf("%d %d\n",A[i],A[i+1]);
	
	
} 
printf("%d %d\n",A[m-1],A[1]+n);
for(i=1;i<m-1;i++){
printf("%d %d\n",A[i]+n,A[i+1]+n);
	
	
} 


 
 
return 0;
}