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:
/*
	构造
	n为2的幂时显然无解 
	发现n大于3并且为奇数时很容易可以构造出一组解 
	每次增加两个时，两两配对即可
	如果最后还剩一个数也可以很容易配出来 
*/
#include <bits/stdc++.h>

using namespace std;

int n;

inline void link(int u, int v) {
	printf("%d %d\n", u, v);
}

int main() {
	scanf("%d", &n);
	if(n < 3 || (n & -n) == n) {
		puts("No");
	} else {
		puts("Yes");
		link(1, 2);
		link(2, 3);
		link(3, 1 + n);
		link(1 + n, 2 + n);
		link(2 + n, 3 + n);
		for(int i = 4; i < n; i += 2) {
			link(1, i);
			link(i, i + 1);
			link(1, i + 1 + n);
			link(i + 1 + n, i + n);
		}
		if(~n & 1) {
			int x = n;
			x ^= 1;
			for(int i = 2; i <= n; ++i) {
				int r = x ^ i;			
				if(r != 1 && r < n) {
					link(n, (i & 1 ? i + n : i));
					link(n + n, (r & 1 ? r + n : r));
					break;
				}
			}
		}
	}
	return 0;
}