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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		br.close();

		int a = 0;
		for (int i = 0; i < 19; i++) {
			a = 1 << i;
			if (n == a) {
				System.out.println("No");
				return;
			}
			if (n < a) {
				a /= 2;
				break;
			}
		}

		System.out.println("Yes");
		System.out.println("1 2");
		System.out.println("2 3");
		System.out.println("3 " + (n + 1));
		System.out.println((n + 1) + " " + (n + 2));
		System.out.println((n + 2) + " " + (n + 3));

		int m = n;
		if (m % 2 == 0) {
			m--;
		}
		for (int i = 4; i < m; i += 2) {
			System.out.println((n + 1) + " " + i);
			System.out.println(i + " " + (i + 1));
			System.out.println((n + 1) + " " + (n + i + 1));
			System.out.println((n + i + 1) + " " + (n + i));
		}

		if (m != n) {
			System.out.println(n + " " + a);
			int b = n - a + 1;
			if (b > 3 && b % 2 == 1) {
				b += n;
			}
			System.out.println(n * 2 + " " + b);
		}
	}
}