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;
bool pt(int n) {
	return (n&(-n)) == n;
}
void eg(int a, int b) {
	cout << a << " " << b << '\n';
}
int lso(int n) {
	return n&(-n);
}
int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	int n;
	cin >> n;
	if(pt(n)) {
		cout << "No\n";return 0;
	}
	if(n%4 != 3 && n <= 4) {
		cout << "No\n";return 0;
	}
	cout << "Yes\n";
	for(int i=0;i<2;i++) {
		eg(i+1,i+2);
	}
	eg(3,n+1);
	for(int i=0;i<2;i++) {
		eg(n+i+1,n+i+2);
	}
	for(int i=4;i<n;i+=2) {
		eg(i,i+1);
		eg(i+1,1);
		eg(1,i+n);
		eg(i+n,i+n+1);
	}
	if(n%2 == 0) {
		int a = lso(n),b = (n-lso(n))+1;
		eg(n,a+n);
		eg(b,n+n);
	}
}