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 Rust (1.15.1) (judge verdict: Wrong Answer)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

#[allow(dead_code)]
fn yn(result: bool) {
    if result {
        println!("Yes");
    } else {
        println!("No");
    }
}

use std::collections::HashSet;
fn main() {
    let n = read::<u32>();
    let mut v = Vec::new();
    let mut hs = HashSet::new();

    for i in 1..n {
        for j in i..n+1 {
            let k = i ^ j;
            if k <= n {
                v.push((i, j, k));
                hs.insert(i);
                hs.insert(j);
                hs.insert(k);
            }
        }
    }

    if hs.len() != n as usize {
        yn(false);
        return;
    }

    for (i, j, k) in v {
        if i == 1 && j == 2 && k == 3 {
            println!("{} {}", i, j);
            println!("{} {}", j, k);
            println!("{} {}", k, i*2);
            println!("{} {}", i*2, j*2);
            println!("{} {}", j*2, k*2);
        }else {
            println!("{} {}", i, j);
            println!("{} {}", j, k);
            println!("{} {}", i, k*2);
            println!("{} {}", i, j*2);
        }
    }
}