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 <cstdio>
#include <algorithm>

using namespace std;

void out(int u, int v){
	printf("%d %d\n", u, v);
}

int main(){
	int n, tmp = 1; scanf("%d", &n);
	while(tmp <= n) tmp <<= 1;
	tmp >>= 1;
	if(tmp == n) return puts("No"), 0;
	puts("Yes");
	for(int i = 2; i < n; i += 2){
		int u = i, v = i + 1;
		out(u, 1); out(v, 1); out(u, v + n), out(v, u + n);
	} out(n + 1, 2 + n);
	if(n % 2 == 0){
		int p = n ^ 1 ^ tmp;
		out(tmp, n + n);
		if(!p) out(1, n);
		else out(p, n);
	}
	return 0;
}