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;

int main()
{
	int n, t = 1<<30;
	scanf("%d", &n);
	while(t > n) t >>= 1;
	if(n == t) puts("No");
	else
	{
		puts("Yes");
		for(int i=2; i<n; i+=2)
		{
			printf("%d %d\n", 1, i);
			printf("%d %d\n", i, i+1);
			printf("%d %d\n", 1, n+i+1);
			printf("%d %d\n", n+i+1, n+i);
		}
		printf("%d %d\n", n+1, 3);
		if(!(n&1))
		{
			printf("%d %d\n", n, n^t);
			printf("%d %d\n", n*2, n+(t^1));
		}
	}
	return 0;
}