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()
{
	cin >> n;
	for (int i = 1; i <= n; i *= 2)
		if (i == n) { cout << "No" << endl; return 0; }
	cout << "Yes" << endl;
	cout << "1 2\n2 3\n";
	cout << 3 << ' ' << n + 1 << endl;
	cout << n + 1 << ' ' << n + 2 << endl;
	cout << n + 2 << ' ' << n + 3 << endl;
	for (int i = 4; i + 1 <= n; i += 2)
	{
		printf("%d %d\n", i, i + 1);
		printf("%d %d\n", n + 1, i + 1);
		printf("%d %d\n", n + i, n + i + 1);
		printf("%d %d\n", n + 1, n + i);
	}
	if (!(n & 1))
	{
		int i = 1;
		while (i * 2 <= n)i *= 2;
		cout << n << ' ' << n+i << endl;
		i ^= n ^ 1;
		cout << n + n << ' ' << i << endl;
	}
	return 0;
}