Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3

Sample Output 1

Yes
1 2
2 3
3 4
4 5
5 6

The sample output represents the following graph:

Sample Input 2

1

Sample Output 2

No

There is no tree satisfying the condition.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>

using namespace std;

int main() {
    int n, i;
    
    scanf("%d", &n);
    
    if (__builtin_popcount(n) == 1) {
        puts("No");
        
        return 0;
    }
    
    puts("Yes");
    
    if (n % 4 == 3) {
        for (i = 1; i < n * 2; i++) printf("%d %d\n", i, i + 1);
    } else if (n % 4 == 1) {
        for (i = 1; i < n - 2; i++) printf("%d %d\n", i, i + 1);
        printf("%d %d\n", n - 2, n + 1);
        for (i = 1; i < n - 2; i++) printf("%d %d\n", i + n, i + n + 1);
        printf("%d %d\n", n - 1, n);
        printf("%d %d\n", n, 1);
        printf("%d %d\n", 1, n * 2 - 1);
        printf("%d %d\n", n * 2 - 1, n * 2);
    } else if (n % 4 == 2) {
        for (i = 1; i < n - 3; i++) printf("%d %d\n", i, i + 1);
        printf("%d %d\n", n - 3, n + 1);
        for (i = 1; i < n - 3; i++) printf("%d %d\n", i + n, i + n + 1);
        printf("%d %d\n", n - 2, n - 1);
        printf("%d %d\n", n - 1, 1);
        printf("%d %d\n", 1, n * 2 - 2);
        printf("%d %d\n", n * 2 - 2, n * 2 - 1);
        printf("%d %d\n", n - 1, n);
        printf("%d %d\n", 2, n * 2);
    } else {
        int x = __builtin_ctz(n) + 1;
        
        for (i = 1; i < n - 1; i++) printf("%d %d\n", i, i + 1);
        printf("%d %d\n", n - 1, n + 1);
        for (i = 1; i < n - 1; i++) printf("%d %d\n", i + n, i + n + 1);
        printf("%d %d\n", n - 1, n);
        printf("%d %d\n", n + (1 << x) - 2, n * 2);
    }
    
    return 0;
}