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 <cstdio>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
int n,m,k;

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