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 Maxn 200007
int n;
void print(int x,int y)
{
    printf("%d %d\n",x,y);
}
int main()
{
    scanf("%d",&n);
    int tmp=n;
    while (tmp%2==0) tmp/=2;
    if (tmp==1)
    {
        printf("No\n");
        return 0;
    }
    printf("Yes\n");
    print(1,2);
    print(2,3);
    print(3,n+1);
    print(n+1,n+2);
    print(n+2,n+3);
    for (int i=4;i<n;i+=2)
    {
        print(i,i+1);
        print(i+1,n+2);
        print(n+3,n+i);
        print(n+i,n+i+1);
    }
    if (n%2==0)
    {
        print(n-1,n);
        int r=(n-1)^n;
        if (r==3) print(n+1,n+n);
        else
        {
            r^=1;
            print(n+r,n+n);
        }
    }
    return 0;
}