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;

int main() {
	int n;
	cin >> n;
	for (int i = 0; i < 20; i++) {
		if (n == 1 << i) {
			cout << "No" << endl;
			return 0;
		}
	}
	cout << "Yes\n";
	for (int i = 2; i < n; i += 2) {
		cout << "1 " << i << '\n';
		cout << i << ' ' << i + 1 << '\n';
		cout << "1 " << i + 1 + n << '\n';
		cout << i + 1 + n << ' ' << i + n << '\n';
	}
	cout << 2 + n << ' ' << 1 + n << endl;
	if (n & 1) return 0;
	int a = 1;
	while (a + a < n) a += a;
	cout << n << ' ' << n - a << '\n';
	cout << n + n << ' ' << a + 1 + n << endl;
	return 0;
}