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;
	ignore = scanf("%d", &n);
	
	if ((n & (n - 1)) == 0) {
		printf("%s\n", "No");
		return 0;
	}
	
	printf("%s\n", "Yes");
	
	int k = n;
	while (k % 4 != 3) k--;
	
	printf("%d %d\n", k, n + 1);
	for (int i = 1; i < k; i++) {
		printf("%d %d\n", i, i + 1);
		printf("%d %d\n", n + i, n + i + 1);
	}
	
	if (n == k + 1) {
		int a = n & (n - 1);
		int b = n ^ a;
		printf("%d %d\n", a + 1, n);
		printf("%d %d\n", n + b, n + n);
	}
	else if (n > k + 1) {
		printf("%d %d\n", 2, k + 1);
		printf("%d %d\n", k + 1, k + 2);
		printf("%d %d\n", 3, n + k + 2);
		printf("%d %d\n", n + k + 2, n + k + 1);
		
		if (n == k + 3) {
			printf("%d %d\n", 2, k + 3);
			printf("%d %d\n", k + 1, n + k + 3);
		}
	}
	
	return 0;
}