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>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define dow(i,j,k) for (int i = j; i >= k; i--)
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
typedef pair<int, int> pi;
typedef long long ll;
int n;
int main() {
	scanf("%d", &n);
	if ((!(n & (n - 1)))) return puts("No"), 0;
	puts("Yes");
	int m = n;
	if (n % 2 == 0) m--;
	printf("%d %d\n", n + 1, 3);
	for (int i = 2; i < m; i += 2) {
		printf("%d %d\n", i, i + 1);
		printf("%d %d\n", i, 1);
		printf("%d %d\n", i + n + 1, i + n);
		printf("%d %d\n", i + n + 1, 1);
	}
	if (n % 2 == 0) {
		int x = n & -n;
		int y = n - x;
		printf("%d %d\n", n, x);
		printf("%d %d\n", y + n + 1, n * 2);
	}
	return 0;
}