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>
int main() {
	int N, M; scanf("%d", &N);
	for(M = 0; (M << 1 | 1) <= N; M = (M << 1 | 1));
	if( N == 1 || M + 1 == N )
		puts("No");
	else {
		puts("Yes");
		int P = (M + 1) >> 1;
		for(int i=1;i<P;i++) {
			printf("%d %d\n", i + N + P, i);
			printf("%d %d\n", i, P);
			printf("%d %d\n", P, i + P);
			printf("%d %d\n", i + P, i + N);
		}
		printf("%d %d\n", P + N, N + 1);
		if( M != N ) {
			printf("%d %d\n", P, M + 1);
			printf("%d %d\n", M + 1 + N, M + 2 + N);
			for(int i=M+2;i<=N;i++) {
				int x = i - (M + 1);
				if( x < P ) printf("%d %d\n", i + N, x + P);
				else if( x == P ) printf("%d %d\n", i + N, x);
				else printf("%d %d\n", i + N, x - P);
				printf("%d %d\n", M + 1, i);
			}
		}
	}
}