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 <stdio.h>

int main() {
	int N;
	scanf("%d", &N);

	int t = 1;
	while(t < N) t *= 2;
	if(t == N) {
		printf("No");
		return 0;
	}
	printf("Yes\n");
	for(int i = 1; i <= (N - 1) / 2; i++) {
		printf("%d %d\n%d 1\n1 %d\n%d %d\n", 2*i, 2*i+1, 2*i+1, 2*i+N, 2*i+N, 2*i+1+N);
	}
	printf("%d 2\n", N+1);
	if(N % 2 == 0) {
		int a = N & (-N);
		int b = N + 1 - a;
		printf("%d %d\n%d %d\n", N, a % 2 ? a : a + N, 2 * N, b % 2 ? b : b + N);
	}
	return 0;
}