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>

const int N = 5e5 + 50;

int n;

int main()
{
	scanf("%d", &n);
	if (n & n - 1)
	{
		printf("Yes\n1 2\n2 3\n3 %d\n%d %d\n%d %d\n", n + 1, n + 1, n + 2, n + 2, n + 3);
		for (int k = 4; k < n; k += 2) printf("%d %d\n%d %d\n%d %d\n%d %d\n", n + 1, k, n + 1, k + 1, k, k + n + 1, k + n, k + 1);
		if ((n & 1) == 0)
		{
			int c = n + 1, j = 2;
			std::function<bool(int)> valid = [&](int u) { return 2 <= u && u <= n; };
			while (!valid(c ^ j)) ++j;
			assert(j <= n);
			if (j == 2) printf("%d %d\n%d %d\n", j + n, n, c ^ j, n << 1);
			else printf("%d %d\n%d %d\n", j, n, c ^ j, n << 1);
		}
	}
	else puts("No");
}