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 <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;

LL LOG[100005];
LL pw2[100005];

void add_edge(LL u,LL v){
	cout << u << ' ' << v << endl;
}

int main(){
	LL n; cin >> n;
	LOG[0] = -1; for(LL i = 1;i <= n;i ++) LOG[i] = LOG[i >> 1] + 1;
	for(LL i = 1;i <= n;i ++) pw2[i] = (1LL << LOG[i]);
	if(n == pw2[n]){
		cout << "No" << endl;
		return 0;
	}
	cout << "Yes" << endl;
	add_edge(1,2); add_edge(2,3); add_edge(3,1 + n); add_edge(1 + n,2 + n); add_edge(2 + n,3 + n);
	for(LL i = 4;i + 1 <= n;i += 2){
		add_edge(i,1); add_edge(i + 1,1);
		add_edge(i + n,i + 1); add_edge(i + n + 1,i);
	}
	if(!(n & 1)){
		add_edge(pw2[n] + 1,n);
		add_edge(n - pw2[n],n + n);
	}
	return 0;
}
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