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 <cstdio>
using namespace std;

namespace zyt
{
	int work()
	{
		int n, id = 0;
		scanf("%d", &n);
		while ((1 << id) <= n)
		{
			if (n == (1 << id))
			{
				puts("No");
				return 0;
			}
			++id;
		}
		--id;
		puts("Yes");
		printf("1 2\n2 3\n3 %d\n%d %d\n%d %d\n", n + 1, n + 1, n + 2, n + 2, n + 3);
		for (int i = 4; i + 1 <= n; i += 2)
		{
			printf("%d %d\n", i, n + 1);
			printf("%d %d\n", i + 1, i);
			printf("%d %d\n", n + 1, i + n + 1);
			printf("%d %d\n", i + n + 1, n + i);
		}
		if (!(n & 1))
		{
			printf("%d %d\n", n, 1 << id);
			printf("%d %d\n", n << 1, n - (1 << id) + 1 == 3 ? 3 : n * 2 - (1 << id) + 1);
		}
		return 0;
	}
}
int main()
{
	return zyt::work();
}