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;

namespace TYC
{
	void work()
	{
		int n, p;
		scanf("%d", &n);
		for (p = 0; (1 << p) <= n; p++)
			if ((1 << p) == n)
				return void(puts("No"));
		puts("Yes");
		p--;
		printf("%d %d\n", 1, 2);
		printf("%d %d\n", 2, 3);
		printf("%d %d\n", 3, n + 1);
		printf("%d %d\n", n + 1, n + 2);
		printf("%d %d\n", n + 2, n + 3);
		for (int i = 4; i < n; i += 2)
		{
			printf("%d %d\n", n + 1, i);
			printf("%d %d\n", i, i + 1);
			printf("%d %d\n", n + 1, n + i + 1);
			printf("%d %d\n", n + i + 1, n + i);
		}
		if (n % 2 == 0)
		{
			printf("%d %d\n", 1 << p, n); 
			if (n - (1 << p) + 1 == 3)
				printf("%d %d\n", 3, n << 1);
			else
				printf("%d %d\n", (n << 1) - (1 << p) + 1, n << 1); 
		}
	}
}

int main()
{
	TYC::work();
	return 0;
}