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;
#define rep(i,n) for(int i = 0; i < n; i++)
int n, t=1, u=2;
int main(){
	scanf("%d", &n);
	for(;t<n;t*=2);
	if(t==n) return 0*puts("No");
	puts("Yes");
	if(n==3) return 0*printf("1 2\n2 3\n3 4\n4 5\n5 6\n");
	t/=2;
	for(;u<t;u++) printf("%d %d\n%d %d\n", u-1,u,u-1+n,u+n);
	printf("%d %d\n", u-1,1+n);
	for(;u<n;u+=2) printf("%d %d\n%d 1\n1 %d\n%d %d\n", u,u+1,u+1,u+n,u+n,u+n+1);
	if(n==u) printf("%d %d\n%d %d\n", n,u%4==2?t+1:t+n,u^t,n+n);
}