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;

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
	for (; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

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

int main()
{
	int n = read();
	for (int i = 0; (1 << i) <= n; ++i) 
		if (n == (1 << i)) return puts("No"), 0;
	
	puts("Yes");
	for (int i = 2; i <= n - 1; i += 2)
	{
		addedge(1, i);
		addedge(i, i + 1);
		addedge(1, i + n + 1);
		addedge(i + n + 1, i + n);
	}
	addedge(3, n + 1);
	
	if (n % 2 == 0)
	{
		int j = 0;
		for (int i = 0; (1 << i) <= n; ++i)
			if (n & (1 << i)) j = i;
		addedge(n, ((1 << j) ^ 1) + n);
		addedge(n + n, n ^ (1 << j));
	}
	return 0;
}