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: Wrong Answer)

use std::io;
//use std::collections::HashMap;
//use std::collections::BinaryHeap;
//use std::collections::HashSet;
//use std::mem; // mem::swap(&mut x,&mut y);
use std::cmp; // cmp::max,cmp::min

// Rust memo: https://hackmd.io/i0lQY1OKTDW1t66TECOmiw?both

#[allow(dead_code)]
fn read_line() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

#[allow(dead_code)]
fn read_charvec() -> Vec<char> {
    read_line().chars().collect()
}

#[allow(dead_code)]
fn read_ints() -> Vec<i64> {
    let s = read_line();
    let split: Vec<&str> = s.split(" ").collect();
    split
        .iter()
        .map(|&x| x.to_string().parse().unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_int() -> i64 {
    read_ints()[0]
}

#[allow(dead_code)]
fn reverse(s: &String) -> String {
    s.chars().rev().collect::<String>()
}

#[allow(dead_code)]
fn read_lines(n: i64) -> Vec<String> {
    let mut xs = Vec::new();

    for _i in 0..n {
        xs.push(read_line());
    }

    xs
}

#[allow(dead_code)]
fn read_ints2(n: i64) -> Vec<i64> {
    read_lines(n).iter().map(|x| x.parse().unwrap()).collect()
}

#[allow(dead_code)]
fn arange(start: i64, end: i64, step: i64) -> Vec<i64> {
    let mut ret = Vec::new();
    let mut i = start;

    while i < end {
        ret.push(i);
        i += step;
    }

    ret
}

#[allow(dead_code)]
const DXS: [i64; 4] = [0, 0, 1, -1];
#[allow(dead_code)]
const DYS: [i64; 4] = [1, -1, 0, 0];
#[allow(dead_code)]
const INF: i64 = 99999999999;
#[allow(dead_code)]
const N_PRIME: i64 = 1000000007;

////////////////////////////////////////////////

fn main() {
    let s = read_charvec();

    if s.len() % 2 == 1 {
        println!("No");
        return;
    }

    for (i, c) in s.into_iter().enumerate() {
        if i % 2 == 0 && c == 'h' {

        } else if i % 2 == 1 && c == 'i' {

        } else {
            println!("No");
            return;
        }
    }

    println!("Yes");
}