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 n;
int ask(int x){
    if(x&1) return x+n;
    return x;
}
int main(){
    scanf("%d",&n);
    if(n<3) return printf("No\n"),0;
    int y=n,flag=0;
    while(y>1){
    flag|=y&1;
    y>>=1;
    }
    if(!flag) return printf("No\n"),0;
    printf("Yes\n");
    printf("%d %d\n",n+1,n+2);
    for(int i=2;i<n;i+=2){
    printf("%d %d\n",1,i);
    printf("%d %d\n",i,i+1);
    printf("%d %d\n",1,i+n+1);
    printf("%d %d\n",i+n,i+n+1);
    }
    if(n%2==0){
    for(int i=2;i<n;i++){
        if(((n^1^i)<n)&&((n^1^i)>1)&&(n^1^i)!=i){
        printf("%d %d\n",ask(i),n);
        printf("%d %d\n",ask(n^1^i),2*n);
        return 0;
        }
    }
    }
}