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>

int n;
int main()
{
	scanf("%d",&n);
	int j = n;
	while((j & 1) == 0) j >>= 1;
	if(j == 1) {
		printf("No\n");
		return 0;
	} else {
		printf("Yes\n");
		printf("%d %d\n", 1, n+2);
		printf("%d %d\n", n+2, 3);
		printf("%d %d\n", 3, n+1);
		printf("%d %d\n", n+1, 2);
		printf("%d %d\n", 2, n+3);
		for(int i = 4;i < n;i += 2) {
			printf("%d %d\n", n+1, i);
			printf("%d %d\n", n+1, i+1);
			printf("%d %d\n", i, n+i+1);
			printf("%d %d\n", i+1, n+i);
		}
		if(n % 2 == 0) {
			for(int j = 2;j < n;++j) if((n^j^1)>=2 && (n^j^1)<n) {
				int a = n^j^1;
				printf("%d %d\n", n, a);
				printf("%d %d\n", 2*n, j);
				break;
			}
		}
	}
}