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;
    cin>>N;
    for(int i=0;i<30;i++){
        if(N==(1<<i)) {printf("No\n"); return 0;}
    }

    printf("Yes\n");
    printf("%d %d\n", N+1, N+3);
    for(int i=3;i<=N;i+=2){
        printf("%d %d\n", 1, i);
        printf("%d %d\n", 1, i-1);
        printf("%d %d\n", i, i+N-1);
        printf("%d %d\n", i-1, i+N);
    }

    if(N%2==0){
        for(int i=2;i<N;i++){
            if(((N+1)^i)<N){
                printf("%d %d\n", i, N);
                printf("%d %d\n", (N+1)^i, 2*N);
                break;
            }
        }
    }
    
    return 0;
}