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>

const int N = 1e5 + 5;

int n, L = 1, tot, a[N];
bool vis[N];

int main()
{
	std::cin >> n; if (n == 1) return puts("No"), 0;
	while ((L << 1) <= n + 1) L <<= 1;
	if (L == n) return puts("No"), 0;
	puts("Yes");
	for (int i = 1; i < L; i <<= 1) vis[a[++tot] = i] = 1;
	for (int i = 1; i < L; i++) if (!vis[i]) a[++tot] = i;
	for (int i = 1; i < tot; i++) printf("%d %d\n", a[i], a[i + 1]);
	printf("%d %d\n", a[tot], a[1] + n);
	for (int i = 1; i < tot; i++) printf("%d %d\n", a[i] + n, a[i + 1] + n);
	for (int i = L; i < n; i += 2)
		printf("%d %d\n%d %d\n%d %d\n%d %d\n",
			i, n + 1, i + 1, i, 1, i + 1 + n, i + 1 + n, i + n);
	if (!(n & 1)) printf("%d %d\n%d %d\n",
		n, n * 2 - 1, n & -n, n * 2);
	return 0;
}