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)

fun main(args: Array<String>) {
    val N = readInputLine().toInt()
    val edges = Array(N + 1) { mutableListOf<Int>() }

    repeat(N - 1) {
        val (a, b) = readInputLine().split(" ").map { it.toInt() }
        edges[a].add(b)
        edges[b].add(a)
    }

    val pairs = Array(N + 1) { mutableListOf<Int>() }
    val usedCnt = mutableMapOf<Int, Int>()

    fun dfs(node: Int, parent: Int, depth2: MutableSet<Int>, depth3: MutableSet<Int>): Pair<MutableSet<Int>, MutableSet<Int>> {
        for (p in depth3) {
            if (parent != 0) {
                pairs[node].add(p)
                pairs[p].add(node)
                usedCnt[node] = (usedCnt[node] ?: 0) + 1
                usedCnt[p] = (usedCnt[p] ?: 0) + 1
            }
        }

        val childs = mutableSetOf<Int>()
        val childDepth1 = mutableSetOf<Int>()
        val childDepth2 = mutableSetOf<Int>()

        for (c in edges[node]) {
            if (c == parent) {
                continue
            }
            val (cd1, cd2) = dfs(
                c,
                node,
                childs.union(childDepth1).toMutableSet(),
                depth2.union(childDepth2).toMutableSet())
            childDepth1.addAll(cd1)
            childDepth2.addAll(cd2)
            childs.add(c)
        }

        return Pair(childs, childDepth1)
    }

    dfs(1, 0, mutableSetOf(), mutableSetOf())

    val ans = IntArray(N)

    val used = mutableSetOf<Int>()

    val sortedCnt = usedCnt.toList().sortedByDescending { it.second }

    var threeIndexCnt = 1
    var nonThreeIndexCnt = 0

    for ((index, cnt) in sortedCnt) {
        if (used.contains(index)) {
            continue
        }
        when {
            cnt > 1 -> {
                if (threeIndexCnt * 3 > N) {
                    println(-1)
                    return
                }
                ans[threeIndexCnt * 3 - 1] = index
                threeIndexCnt++
                used.add(index)
            }
            cnt == 1 -> {
                val pair = pairs[index].first()
                if (!used.contains(pair)) {
                    if (nonThreeIndexCnt / 2 * 6 + nonThreeIndexCnt % 2 + 3 < N) {
                        ans[nonThreeIndexCnt / 2 * 6 + nonThreeIndexCnt % 2] = index
                        ans[nonThreeIndexCnt / 2 * 6 + nonThreeIndexCnt % 2 + 3] = pair
                        nonThreeIndexCnt++
                    } else {
                        ans[threeIndexCnt * 3 - 1] = index
                        ans[(threeIndexCnt + 1) * 3 - 1] = pair
                        threeIndexCnt += 2
                    }
                    used.add(index)
                    used.add(pair)
                }
            }
        }
    }

    var currentIndex = 0

    for (i in 1..N) {
        if (used.contains(i)) {
            continue
        }
        while (ans[currentIndex] != 0) {
            currentIndex++
        }
        ans[currentIndex] = i
    }

    println(ans.joinToString(" "))
}

fun readInputLine(): String {
    return readLine()!!
}