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 ll long long
using namespace std;
int main()
{
	int n,i,j,p;cin>>n;
	for (i=0;i<=20;i++) if ((1<<i)==n){puts("No");exit(0);}else if ((1<<i)<n) p=i;
	puts("Yes");
	if (n&1)
	{
		for (i=2;i<=n;i+=2)
		{
			cout<<i<<' '<<i+1<<endl;
			cout<<i+n<<' '<<i+n+1<<endl;
			cout<<1<<' '<<i+1<<endl;
			cout<<1<<' '<<i+n<<endl;
		}
		cout<<n+1<<' '<<n+3;
	}
	else
	{
		int a=(1<<p),b=n-a;
		for (i=2;i<n;i+=2) 
		{
			cout<<i<<' '<<i+1<<endl;
			cout<<i+n<<' '<<i+n+1<<endl;
			if (i!=a) cout<<1<<' '<<i+1<<endl,cout<<1<<' '<<i+n<<endl;
		}
		cout<<b<<' '<<a<<endl;
		cout<<b+n+1<<' '<<a+n+1<<endl;
		cout<<n<<' '<<a<<endl;
		cout<<n+n<<' '<<b<<endl;
		cout<<n+1<<' '<<b+n+1;
	}
}
//a+1 a b b+1 1 b+n b+n+1 a+n+1 a+n
//5 4 2 3 1 8 9 11 10