#include <cstdio>

int main() {
	int N;
	scanf("%d", &N);
	if (!(N & (N - 1)))
		return puts("No"), 0;
	puts("Yes");
	printf("%d %d\n", N + 1, 3);
	for (int i = 2; i < N; i += 2)
		printf("%d %d\n%d %d\n%d %d\n%d %d\n", i + 1, i, i, 1, 1, N + i + 1, N + i + 1, N + i);
	if (N % 2 == 0)
		for (int i = 2; i < N; ++i)
			if ((i ^ N ^ 1) >= 2 && (i ^ N ^ 1) < N) {
				printf("%d %d\n%d %d\n", i % 2 == 0 ? i : N + i, N, (i ^ N ^ 1) % 2 == 0 ? i ^ N ^ 1 : N + (i ^ N ^ 1), 2 * N);
				break;
			}
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <cstdio>

int main() {
	int N;
	scanf("%d", &N);
	if (!(N & (N - 1)))
		return puts("No"), 0;
	puts("Yes");
	printf("%d %d\n", N + 1, 3);
	for (int i = 2; i < N; i += 2)
		printf("%d %d\n%d %d\n%d %d\n%d %d\n", i + 1, i, i, 1, 1, N + i + 1, N + i + 1, N + i);
	if (N % 2 == 0)
		for (int i = 2; i < N; ++i)
			if ((i ^ N ^ 1) >= 2 && (i ^ N ^ 1) < N) {
				printf("%d %d\n%d %d\n", i % 2 == 0 ? i : N + i, N, (i ^ N ^ 1) % 2 == 0 ? i ^ N ^ 1 : N + (i ^ N ^ 1), 2 * N);
				break;
			}
	return 0;
}

```