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() {
	scanf("%d", &n);
	if (n == (n & -n)){
		printf("No"); return 0;
	}
	puts("Yes");
	printf("%d %d\n", 1, 2);
	printf("%d %d\n", 2, 3);
	printf("%d %d\n", 3, n + 1);
	printf("%d %d\n", n + 1, n + 2);
	printf("%d %d\n", n + 2, n + 3);
	for (int i = 4; i + 1 <= n; i += 2) {
		printf("%d %d\n", i, i + 1);
		printf("%d %d\n", i + n, i + n + 1);
		printf("%d %d\n", n + 1, i + 1);
		printf("%d %d\n", n + 1, i + n);
	}
	if (~n & 1) {
		for (int i = 2; i < n; ++i) {
			int x = i, y = n ^ i ^ 1;
			if (y >= 2 && y < n && x != y) {
				printf("%d %d\n", n, x & 1 ? x : x + n);
				printf("%d %d\n", n + n, y & 1 ? y : y + n);
				break;
			}
		}
	}
}