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;
typedef long long ll;
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll n;cin>>n;
    if(__builtin_popcount(n)==1)return cout<<"No" , 0;
    cout<<"Yes\n"<<"2 "<<n+1<<"\n";
    for(ll x=2;x<n;x+=2){
        cout<<x<<" "<<x+1<<"\n";
        cout<<"1 "<<x+1<<"\n";
        cout<<"1 "<<x+n<<"\n";
        cout<<x+n<<" "<<x+n+1<<"\n";
    }

    ll k=0;while((1ll<<k)<n)k++;
    k--;

    if(n%2==0){
        cout<<n<<" "<<n+(1ll<<k)<<"\n";
        cout<<2*n<<" "<<(n-(1ll<<k))+1<<"\n";
    }
}