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>

int main(int argc, char ** argv) {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	int n; std::cin >> n;
	if (n < 3 || __builtin_popcount(n) == 1)
		return std::cout << "No" << std::endl, 0;
	std::cout << "Yes" << std::endl;
	for (int i = 2; i + 1 <= n; i += 2) {
		std::cout << i << ' ' << i + 1 << '\n';
		std::cout << i + 1 << ' ' << 1 << '\n';
		std::cout << 1 << ' ' << i + n << '\n';
		std::cout << i + n << ' ' << i + n + 1 << '\n';
	}
	std::cout << n + 1 << ' ' << 2 << '\n';
	if (n % 2 == 0) {
		for (int i = 2; i < n; ++i) {
			int t = n ^ 1 ^ i;
			if (2 <= t && t < n) {
				std::cout << n << ' ' << t + n * (t % 2 == 0) << '\n';
				std::cout << n * 2 << ' ' << i + n * (i % 2 == 0) << '\n';
				return 0;
			}
		}
		assert(0);
	}
	return 0;
}