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 n;

void link(int x, int y) { std::cout << x << ' ' << y << '\n'; }

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n;
	if (!(n & n - 1))
		return std::cout << "No\n", 0;
	std::cout << "Yes\n";
	link(1, 3), link(3, 2), link(2, n + 1), link(n + 1, n + 3), link(n + 3, n + 2);
	for (int i = 4; i < n; i += 2)
		link(1, i + 1), link(i + 1, i), link(1, i + n), link(i + n, i + n + 1);
	if (!(n & 1)) {
		int p = 1 << 31 - __builtin_clz(n);
		link(n, p + n), link(n + n, n - p + 1);
	}
	return 0;
}