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;

void p (int x, int y) {
	printf("%d %d\n", x, y);
	
}

int main (void) {
	int n;
	scanf("%d", &n);
	if (n == 3) {
		puts("Yes");
		p(1, 2);
		p(2, 3);
		p(3, 4);
		p(4, 5);
		p(5, 6);
	} else if (n <= 2) {
		puts("No");
	} else if (n & 1) {
		puts("Yes");
		p(1, 2);
		p(2, 3);
		p(3, n + 1);
		p(n + 1, n + 2);
		p(n + 2, n + 3);
		for (int i = 4; i <= n; i += 2) {
			p(i, i + 1);
			p(i + 1, 1);
			p(1, i + n);
			p(i + n, i + n + 1);
		}
	} else {
		int m = (n) & (-n);
		if (m == n) {
			puts("No");
			return 0;
		} 
		puts("Yes");
		p(n + n, m + n);
		p(n, (n - m) ^ 1);
		p(1, 2);
		p(2, 3);
		p(3, n + 1);
		p(n + 1, n + 2);
		p(n + 2, n + 3);
		for (int i = 4; i < n; i += 2) {
			p(i, i + 1);
			p(i + 1, 1);
			p(1, i + n);
			p(i + n, i + n + 1);
		}
	}
}