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)

use std::collections::HashSet;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn dfs(now: usize, pre: usize, nei: &[Vec<usize>], a: &mut Vec<usize>, b: &mut Vec<usize>) {
    a.push(now);
    for &n in nei[now].iter() {
        if n == pre {
            continue;
        }
        dfs(n, now, nei, b, a);
    }
}

fn main() {
    input! {
        n: usize,
        ab: [(usize1, usize1); n - 1],
    };
    let nei = {
        let mut nei = vec![vec![]; n];
        for (a, b) in ab {
            nei[a].push(b);
            nei[b].push(a);
        }
        nei
    };
    let mut a = Vec::new();
    let mut b = Vec::new();
    dfs(0, 0, &nei, &mut a, &mut b);

    let zero_num = n / 3;
    let mut ans = vec![0; n];
    if b.len() < a.len() {
        std::mem::swap(&mut a, &mut b);
    }
    if a.len() <= zero_num {
        let mut used = HashSet::with_capacity(a.len());
        for (i, a) in a.into_iter().enumerate() {
            let j = (i + 1) * 3;
            ans[a] = j;
            used.insert(j);
        }
        for i in 1..n + 1 {
            if used.contains(&i) {
                continue;
            }
            ans[b.pop().unwrap()] = i;
        }
    } else {
        for i in 0..(n + 2) / 3 {
            ans[a.pop().unwrap()] = i * 3 + 1;
        }
        for i in 0..(n + 1) / 3 {
            ans[b.pop().unwrap()] = i * 3 + 2;
        }
        let mut i = 3;
        for a in a {
            ans[a] = i;
            i += 3;
        }
        for b in b {
            ans[b] = i;
            i += 3;
        }
    }
    for a in ans {
        print!("{} ", a);
    }
    println!();
}