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>
#define lowbit(x) ((x)&(-(x)))
using namespace std;

int n;

int main()
{
	cin >> n;
	if(lowbit(n) == n){
		puts("No");
		return 0;
	}
	puts("Yes");
	for(int i = 2; i < n; i += 2){
		printf("1 %d\n", i), printf("%d %d\n", i, i+1);
		printf("1 %d\n", i+n+1), printf("%d %d\n", i+n+1, i+n);
	}
	printf("3 %d\n", n+1);
	if(n&1) return 0;
	int p = lowbit(n);
	printf("%d %d\n", p+n+1, n);
	printf("%d %d\n", n^p, n+n);
	
	return 0;
}