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 pii pair<int,int>
vector<pii> e;
void addedge(int a,int b)
{
    e.emplace_back(a,b);
}
int main()
{
    int n;scanf("%d",&n);
    if(n<3||(n-(n&-n))==0)printf("No\n");
    else{
        printf("Yes\n");
        addedge(1,2),addedge(2,3),addedge(3,n+1);
        addedge(1,3+n),addedge(3+n,2+n);
        for(int i=4;i+1<=n;i+=2)
            addedge(1,i),addedge(i,i+1),addedge(1,i+1+n),addedge(i+1+n,i+n);
        if(!(n&1))
        {
            int k=(n-1)^n;
            addedge(k+n,n),addedge(n-2,n*2);
        }
        for(pii p:e)printf("%d %d\n",p.first,p.second);
    }
}