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 n;

int main() {
	cin >> n;
	int b = 1;
	while (b*2 <= n) b *= 2;
	if (n == b) {
		cout << "No\n";
		return 0;
	}
	cout << "Yes\n";
	cout << "1 2\n";
	cout << "2 3\n";
	cout << "3 " << n+1 << "\n";
	cout << n+3 << " " << 1 << "\n";
	cout << n+2 << " " << n+3 << "\n";
	for (int k = 4; k+1 <= n; k += 2) {
		cout << "1 " << k << "\n";
		cout << k << " " << k+1 << "\n";
		cout << "1 " << n+k+1 << "\n";
		cout << n+k+1 << " " << n+k << "\n";
	}
	if (n%2 == 0) {
		int x = 1^n^b;
		int y = 1;
		cout << x+n << " " << n << "\n";
		cout << b << " " << 2*n << "\n";
	}
}