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 <iostream>

using namespace std;
typedef long long ll;

void out(ll i, ll j){ cout << i << " " << j << endl; }

int main(){
    ll n;
    cin >> n;

    if((n & (n - 1)) == 0){
        cout << "No" << endl;
        return 0;
    }
    cout << "Yes" << endl;

    ll n2 = n - (1 - n % 2);
    out(1, n + 2);
    out(n + 2, n + 3);
    out(n + 3, n + 1);
    out(n + 1, 2);
    out(2, 3);
    for(ll i = 3; i < n2; i+=2){
        out(n + 1, i + 1);
        out(i + 1, i + 2);
        out(n + 1, n + i + 2);
        out(n + i + 2, n + i + 1);
    }
    if(n % 2 == 0){
        ll x = n ^ (n - 1) ^ 1;
        out(n + n - 1, n + n);
        out(x, n);
    }
    return 0;
}