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;
int main() {
    int n; cin >> n;
    if (n < 3 || (n & (n-1)) == 0) { cout << "No\n"; return 0; }
    cout << "Yes\n1 2\n2 3\n";
    cout << 3 << ' ' << 1+n << '\n'
         << 1+n << ' ' << 2+n << '\n'
         << 2+n << ' ' << 3+n << '\n';
    for (int k = 4; k+1 <= n; k+=2)
        cout << k << ' ' << k+1 << '\n'
             << k+1 << ' ' << 1 << '\n'
             << 1 << ' ' << k+n << '\n'
             << k+n << ' ' << k+n+1 << '\n';
    if (n%2==0) {
        int a = n & (n-1), b = n - a;
        cout << a+1 << ' ' << n << '\n'
             << b+n << ' ' << 2*n << '\n';
    }
}