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>
#define add(x, y) printf("%d %d\n", x, y);
int main()
{
	int n;
	scanf("%d", &n);
	if (__builtin_popcount(n) == 1)
	{
		puts("No");
		return 0;
	}
	puts("Yes");
	add(1, 2);
	add(2, 3);
	add(n + 3, 1);
	add(n + 3, n + 2);
	add(n + 2, n + 1);
	for (int i = 4; i + 1 <= n; i += 2)
	{
		add(1, i);
		add(i, i + 1);
		add(1, n + i + 1);
		add(n + i + 1, n + i);
	}
	if (n & 1 ^ 1)
	{
		for (int i = 2; i < n; i++)
		{
			int j = n ^ 1 ^ i;
			if (j < n && j >= 2 && i != j)
			{
				add(i + (i & 1) * n, n);
				add(j + (j & 1) * n, n * 2);
				return 0;
			}
		}
	}
	return 0;
}