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 lowbit(int x) {return x&-x;}
int n;
void work() {
    for(int i=2;i+1<=n;i+=2)
	{
	    cout<<i+n+1<<' '<<i<<'\n';
	    cout<<i<<' '<<1<<'\n';
	    cout<<1<<' '<<i+1<<'\n';
	    cout<<i+1<<' '<<i+n<<'\n';
	}
    cout<<n+1<<' '<<n+2<<'\n';
    if(!(n&1))
	{
	    int x=1;
	    while(x+x<=n) x<<=1;
	    cout<<(x+1)<<' '<<n<<'\n';
	    cout<<(x^n)<<' '<<n+n<<'\n';
	}
}
int main()
{
    cin>>n;
    if(lowbit(n)==n) cout<<"No"<<'\n';
    else
	{
	    cout<<"Yes"<<'\n';
	    work();
	}
    return 0;
}