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;

#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=100005;
int n;
void print(int u,int v){printf("%d %d\n",u,v);}

int main(){
	scanf("%d",&n);
	int t=1;
	while(t<n)t<<=1;
	if(t==n){
		puts("No");
		return 0;
	}
	puts("Yes");
	print(1,2);
	print(2,3);
	print(3,1+n);
	print(1+n,2+n);
	print(2+n,3+n);
	for(int i=4;i<n;i+=2){
		print(i,i+1);
		print(i+1,1);
		print(1,i+n);
		print(i+n,i+1+n);
	}
	if(n%2==0){
		print(n-1,n);
		print((n^(n-1))-1+n,n+n);
	}
	return 0;
}