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 n;

int main()
{
	int i, t;
	
	scanf("%d", &n);
	
	if(n < 3 || ((~n & 1) && (n & n - 1) == 0)){
		printf("No\n");
		return 0;
	}
	
	printf("Yes\n");
	
	if(n & 1){
		printf("1 2\n2 3\n3 %d\n%d %d\n%d %d\n", n + 1, n + 1, n + 2, n + 2, n + 3);
		
		for(i=4; i<n; i+=2){
			printf("1 %d\n%d %d\n1 %d\n%d %d\n", i, i, i + 1, n + i + 1, n + i + 1, n + i);
		}
	}
	else{
	}
	
	if(~n & 1){
		t = (n & -n);
		
		for(i=2; i<n; i+=2){
			if(i == t){
				printf("1 %d\n%d %d\n%d %d\n%d %d\n%d %d\n", i, i, i + 1, i + 1, n + 1, n + 1, n + i, n + i, n + i + 1);
			}
			else{
				printf("1 %d\n%d %d\n1 %d\n%d %d\n", i, i, i + 1, n + i + 1, n + i + 1, n + i);
			}
		}
		
		printf("%d %d\n%d %d\n", n, n + n - t + 1, t, n + n);
	}
	
	return 0;
}