Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices. The vertices are numbered 1 to N, and the i-th edge connects Vertex a_i and Vertex b_i.

Takahashi loves the number 3. He is seeking a permutation p_1, p_2, \ldots , p_N of integers from 1 to N satisfying the following condition:

For every pair of vertices (i, j), if the distance between Vertex i and Vertex j is 3, the sum or product of p_i and p_j is a multiple of 3.

Here the distance between Vertex i and Vertex j is the number of edges contained in the shortest path from Vertex i to Vertex j.

Help Takahashi by finding a permutation that satisfies the condition.

Constraints

2\leq N\leq 2\times 10^5

1\leq a_i, b_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
\vdots
a_{N-1} b_{N-1}

Output

If no permutation satisfies the condition, print -1.

Otherwise, print a permutation satisfying the condition, with space in between.
If there are multiple solutions, you can print any of them.

Sample Input 1

5
1 2
1 3
3 4
3 5

Sample Output 1

1 2 5 4 3

The distance between two vertices is 3 for the two pairs (2, 4) and (2, 5).

p_2 + p_4 = 6

p_2\times p_5 = 6

Thus, this permutation satisfies the condition.

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let mut solver = Solver::new(n);
    for _ in 0..n - 1 {
        let a: usize = scan(tokens);
        let b: usize = scan(tokens);
        solver.add_edge(a - 1, b - 1);
    }
    solver.solve();
}

struct Solver {
    n: usize,
    neighbors: Vec<Vec<usize>>,
    color: Vec<bool>,
}

impl Solver {
    fn new(n: usize) -> Solver {
        Solver {
            n: n,
            neighbors: vec![Vec::new(); n],
            color: vec![false; n],
        }
    }
    fn add_edge(&mut self, i: usize, j: usize) {
        self.neighbors[i].push(j);
        self.neighbors[j].push(i);
    }
    fn solve(&mut self) {
        self.give_color(0, 0, false);
        let n = self.n;
        let nt = self.color.iter().filter(|&&c| c).count();
        let nf = n - nt;
        let (nr, r) = if nt <= nf { (nt, true) } else { (nf, false) };
        let n3 = n / 3;
        let mut res = vec![0; n];
        let mut p1 = 1;
        let mut p2 = 2;
        let mut p3 = 3;
        if nr <= n3 {
            for i in 0..n {
                if self.color[i] == r {
                    res[i] = p3;
                    p3 += 3;
                } else if p1 <= n {
                    res[i] = p1;
                    p1 += 3;
                } else if p2 <= n {
                    res[i] = p2;
                    p2 += 3;
                } else {
                    res[i] = p3;
                    p3 += 3;
                }
            }
        } else {
            for i in 0..n {
                if self.color[i] == r {
                    if p1 <= n {
                        res[i] = p1;
                        p1 += 3;
                    } else {
                        res[i] = p3;
                        p3 += 3;
                    }
                } else {
                    if p2 <= n {
                        res[i] = p2;
                        p2 += 3;
                    } else {
                        res[i] = p3;
                        p3 += 3;
                    }
                }
            }
        }
        for i in 0..n {
            if i != 0 {
                print!(" ");
            }
            print!("{}", res[i]);
        }
        println!("");
    }
    fn give_color(&mut self, i: usize, p: usize, c: bool) {
        self.color[i] = c;
        for j in self.neighbors[i].clone() {
            if j != p {
                self.give_color(j, i, !c);
            }
        }
    }
}