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 <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
using namespace std;
int n,k;
void ot(int u,int v){printf("%d %d\n",u,v);;}
int main(){
    scanf("%d",&n);while((1<<k+1)<=n) ++k;
    int m=1<<k;if(m==n) return puts("No"),0;puts("Yes");
    ot(1,3);ot(3,2);ot(2,n+1);ot(n+1,n+3);ot(n+3,n+2);
    rep(i,2,(n-1)/2) ot(1,2*i),ot(1,2*i+1),ot(2*i,2*i+1+n),ot(2*i+1,2*i+n);
    if(!(n&1)){
        int t=m^1^n;
        ot(m,n),ot(t,2*n);
    }
    return 0;
}