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 ref(i,x,y)for(int i=x;i<=y;++i)
#define def(i,x,y)for(int i=x;i>=y;--i)
#define pb push_back
#define SZ(x) ((int)x.size())
#define fi first
#define se second
#define mp make_pair
const int N=100010;
int n, mk[N*3];
int main(){
	cin>>n;
	ref(i,0,17)if((1<<i)==n){
		cout<<"No"<<endl;
		return 0;
	}
    cout<<"Yes"<<endl;
	cout<<1+n<<" "<<3<<endl;
	for(int i=2;i<n;i+=2){
		cout<<1<<" "<<i<<endl; mk[i]=i;
		cout<<i<<" "<<i+1<<endl;
		cout<<1<<" "<<i+1+n<<endl; mk[i+1]=i+1+n;
		cout<<i+1+n<<" "<<i+n<<endl;
	}
	if(n%2==0){
		int m=0;
		while(!(n>>m&1))m++;
		m=1<<m;
		cout<<mk[(n-m)]<<" "<<n<<endl;
		cout<<mk[m+1]<<" "<<n+n<<endl;
	}
}