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)

//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let 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 run() {
    input! {
        n: usize,
        e: [(usize1, usize1); n - 1],
    }
    let mut g = vec![vec![]; n];
    for (a, b) in e {
        g[a].push(b);
        g[b].push(a);
    }
    let mut depth = vec![0; n];
    let mut cnt = vec![0; 2];
    let mut stack = vec![(0, 0)];
    while let Some((v, p)) = stack.pop() {
        cnt[depth[v]] += 1;
        for &u in g[v].iter() {
            if u != p {
                depth[u] = depth[v] ^ 1;
                stack.push((u, v));
            }
        }
    }
    let mut c = vec![0; 3];
    for i in 1..(n + 1) {
        c[i % 3] += 1;
    }
    let one = c[1];
    let two = c[2];
    let mut ans = vec![0; n];
    if depth[0] >= one + two {
        let mut x = 1;
        let mut y = 3;
        for i in 0..n {
            if depth[i] == 0 {
                if x <= n {
                    ans[i] = x;
                    x += 1;
                    if x % 3 == 0 {
                        x += 1;
                    }
                } else {
                    ans[i] = y;
                    y += 3;
                }
            } else {
                ans[i] = y;
                y += 3;
            }
        }
    } else if depth[1] >= one + two {
        let mut x = 1;
        let mut y = 3;
        for i in 0..n {
            if depth[i] == 1 {
                if x <= n {
                    ans[i] = x;
                    x += 1;
                    if x % 3 == 0 {
                        x += 1;
                    }
                } else {
                    ans[i] = y;
                    y += 3;
                }
            } else {
                ans[i] = y;
                y += 3;
            }
        }
    } else if depth[0] >= one && depth[1] >= two {
        let mut x = 1;
        let mut y = 2;
        let mut z = 3;
        for i in 0..n {
            if depth[i] == 0 {
                if x <= n {
                    ans[i] = x;
                    x += 3;
                } else {
                    ans[i] = z;
                    z += 3;
                }
            } else {
                if y <= n {
                    ans[i] = y;
                    y += 3;
                } else {
                    ans[i] = z;
                    z += 3;
                }
            }
        }
    } else {
        let mut x = 1;
        let mut y = 2;
        let mut z = 3;
        for i in 0..n {
            if depth[i] == 1 {
                if x <= n {
                    ans[i] = x;
                    x += 3;
                } else {
                    ans[i] = z;
                    z += 3;
                }
            } else {
                if y <= n {
                    ans[i] = y;
                    y += 3;
                } else {
                    ans[i] = z;
                    z += 3;
                }
            }
        }
    }
    let mut s = String::new();
    for a in ans {
        s.push_str(&format!("{} ", a));
    }
    s.pop();
    println!("{}", s);
}

fn main() {
    run();
}