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;
const int N=1e5+100;
int n;
int main() {
	scanf("%d",&n);
	if(n == (n&(-n)) ) {
		puts("No");
		return 0;
	}
	printf("Yes\n1 2\n2 3\n3 %d\n%d %d\n%d %d\n",n+1,n+1,n+2,n+2,n+3);
	if(n == 3) return 0;
	int cur=4;
	for(;cur<=n-1;cur+=2) 
		printf("%d %d\n%d %d\n%d %d\n%d %d\n",cur,cur+1,cur+1,n+1,n+1,n+cur,n+cur,n+cur+1);
	if(cur == n+1) return 0;
	for(int i=2;i<=n;++i)
		if((n^i^1)<=n) {
			if(i&1) printf("%d %d\n%d %d\n",n,i,n+n,(n^i^1)+n);
			else printf("%d %d\n%d %d\n",n,i+n,n+n,(n^i^1));
			return 0;
		}
	return 0;
}