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 a[1919810];

void P(int u,int v) {
	cout<<u<<' '<<v<<'\n';
}

int main() {
	int n,x=1;cin>>n;
	while(x<n) x<<=1;
	if(n==x) return puts("No"),0;
	x>>=1;
	puts("Yes");
	if(n%4==3) {
		for(int i=1;i<2*n;i++) P(i,i+1);
	}
	else if(n%4==1) {
		for(int i=2;i<n;i++) P(i,i+1),P(i+n,i+n+1);
		P(1,2);P(n,n+2);P(n+1,n+3);
	}
	else if(n%4==0) {
		for(int i=1;i<n;i++) a[i]=i;
		swap(a[x-1],a[n^x]);
		for(int i=1;i<=n-2;i++) P(a[i],a[i+1]),P(n+a[i],n+a[i+1]);
		P(a[n-1],n+a[1]);P(n,x);P(2*n,n^x);
	}
	else {
		for(int i=2;i<n;i++) a[i]=i;
		swap(a[x-1],a[n^x]);
		if((n^x)==2&&n!=6) swap(a[x-2],a[3]);
		for(int i=2;i<=n-2;i++) P(a[i],a[i+1]),P(n+a[i],n+a[i+1]);
		P(a[n-1],n+a[2]);P(n,x);P(2*n,n^x);P(1,2);P(n+1,3);
	}
	return 0;
}