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 OCaml (4.02.3) (judge verdict: Wrong Answer)

let () =
    let main () =
        let n = int_of_string (read_line ()) in
        if n <> 1 && n mod 2 = 1 then (
            print_endline "Yes";
            for i = 1 to (n - 1) / 2 do
                Printf.printf "%d %d\n" 1 (2 * i);
                Printf.printf "%d %d\n" (2 * i) (2 * i + 1);
                Printf.printf "%d %d\n" 1 (2 * i + 1 + n);
                Printf.printf "%d %d\n" (2 * i + n) (2 * i + 1 + n);
            done;
            Printf.printf "%d %d\n" 3 (n + 1)
        ) else if n > 3 && (n - 2) land (n - 3) = 0 then (
            print_endline "Yes";
            for i = 1 to (n - 4) / 2 do
                Printf.printf "%d %d\n" 1 (2 * i);
                Printf.printf "%d %d\n" (2 * i) (2 * i + 1);
                Printf.printf "%d %d\n" 1 (2 * i + 1 + n);
                Printf.printf "%d %d\n" (2 * i + n) (2 * i + 1 + n);
            done;
            Printf.printf "%d %d\n" 3 (n + 1);
            Printf.printf "%d %d\n" 3 (n + n - 1);
            Printf.printf "%d %d\n" (n + 2) (n - 2);
            Printf.printf "%d %d\n" (n - 2) (n - 1);
            Printf.printf "%d %d\n" (n - 2) n;
            Printf.printf "%d %d\n" (n + 2) (n + n);
            Printf.printf "%d %d\n" (n + n) (n + n - 2);
        ) else print_endline "No"
    in
    main ()