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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.ArrayList

private fun readLn() = readLine()!! // string line
private fun readInt() = readLn().toInt() // single int
private fun readLong() = readLn().toLong() // single int
private fun readStrings() = readLn().split(" ") // list of strings
private fun readInts() = readStrings().map { it.toInt() } // list of ints

fun dfs(u: Int, p: Int, lvl: Int, adj: Array<ArrayList<Int>>, V1: ArrayList<Int>, V2: ArrayList<Int>) {
    if (lvl % 2 == 0) {
        V1.add(u)
    } else {
        V2.add(u)
    }
    for (v in adj[u]) {
        if (v == p) continue
        dfs(v, u, lvl + 1, adj, V1, V2)
    }
}

fun main(args: Array<String>) {

    val N = readInt()
    val adj = Array(N) { ArrayList<Int>() }
    for ((u, v) in (0 until N - 1).map { readInts().map { it - 1 } }) {
        adj[u].add(v)
        adj[v].add(u)
    }
    val V1 = ArrayList<Int>()
    val V2 = ArrayList<Int>()
    dfs(0, -1, 0, adj, V1, V2)
    val N1 = (N + 2) / 3
    val N2 = (N + 1) / 3
    val N3 = N / 3
    if (Math.min(V1.size, V2.size) < N2) {
        println(-1)
        return
    }
    val ans = Array(N) { 0 }
    for (i in (1..N).reversed()) {
        if (i % 3 == 1) {
            ans[ V1.last() ] = i
            V1.removeAt(V1.size - 1)
        } else if (i % 3 == 2) {
            ans[ V2.last() ] = i
            V2.removeAt(V2.size - 1)
        } else if (V1.size > V2.size) {
            ans[ V1.last() ] = i
            V1.removeAt(V1.size - 1)
        } else {
            ans[ V2.last() ] = i
            V2.removeAt(V2.size - 1)
        }
    }
    println(ans.joinToString(" "))



}

/*
1 1 5 -> 3
1 5 1 -> 2
5 1 1 -> 1
 */