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<cstdio>
#include<algorithm>
#include<vector>
#define pii pair<int,int>
using namespace std;
int n, D[101000], Old[101000];
vector<pii>OO[101000];
void Add(int a, int b) {
	printf("%d %d\n", a, b);
}
int main() {
	int i, j, k;
	scanf("%d", &n);
	int sz = 1;
	while (sz < n)sz *= 2;
	if (sz == n) {
		puts("No");
		return 0;
	}
	puts("Yes");
	Add(1, 2);
	Add(2, 3);
	Add(3, n+1);
	Add(n+1, n+2);
	Add(n+2, n+3);

	for (i = 4; i < n; i += 2) {
		Add(i, i + 1);
		Add(i + 1, n + 1);
		Add(n + 1, n + i);
		Add(n + i, n + i + 1);
	}
	if (n % 2 == 0) {
		Add(n - 1, n);
		Add(n + (n ^ (n - 1) ^ 1), n + n);
	}
	
}