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 main(){
	int N, k, i;
	scanf("%d", &N);
	for(k = 1; k < N; k <<= 1);
	if(k == N){
		printf("No\n");
		return 0;
	}
	printf("Yes\n");
	k >>= 1;
	if(N == 3){
		printf("1 2\n");
		printf("2 3\n");
		printf("3 4\n");
		printf("4 5\n");
		printf("5 6\n");
		return 0;
	}
	vector<int> a(N + 1);
	for(i = 1; i < k; i++){
		a[i] = i;
	}
	swap(a[2], a[3]);
	for(i = 1; i < k - 1; i++){
		printf("%d %d\n", a[i], a[i + 1]);
		printf("%d %d\n", a[i] + N, a[i + 1] + N);
	}
	printf("%d %d\n", a[k - 1], 1 + N);
	for(i = N; i > k; i -= 2){
		int x = i, y = i - 1;
		printf("%d %d\n", x, y);
		printf("%d %d\n", x + N, y + N);
		printf("%d %d\n", x, x ^ y);
		printf("%d %d\n", y + N, x ^ y);
	}
	if(N % 2 == 0){
		printf("%d %d\n", 1, k);
		printf("%d %d\n", k + 2, k + N);
	}
	return 0;
}