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() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int N;
	cin >> N;
	if (N == (N & -N)) return cout << "No\n", 0;
	auto edge = [] (int u, int v) { cout << u << ' ' << v << '\n'; };
	cout << "Yes\n" << N + 1 << ' ' << N + 2 << '\n';
	for (int u = 2; u < N; u += 2) {
		edge(1, u);
		edge(1, u ^ 1);
		edge(u, N + (u ^ 1));
		edge(u ^ 1, N + u);
	}
	if (!(N & 1)) {
		int lowbit = N & -N;
		edge(N, N ^ lowbit);
		edge(N << 1, 1 ^ lowbit);
	}
}