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 main(){
    int n,n1,n2,m=1,i;
    scanf("%d",&n);
    while((m<<1)<=n)m<<=1;
    if(n==m){printf("No");return 0;}
    n1=n+1;n2=n<<1;
    printf("Yes\n1 2\n2 3\n3 %d\n%d %d\n%d %d\n",n1,n1,n+2,n+2,n+3);
    for(i=4;i<n;i+=2)printf("%d %d\n%d %d\n%d %d\n%d %d\n",n1,i,i,i|1,n1,n+(i|1),n+(i|1),n+i);
    if(!(n&1)){
        printf("%d %d\n",m,n);
        if(n==m+2)printf("3 %d\n",n2);
        else printf("%d %d\n",n2-m+1,n2);
    }
    return 0;
}