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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

int main() {
    int a = 0, b = 0, m = 0, x = 0, y = 0, c = 0;
    int min = 0, amin = 0, bmin = 0;
    int arra[100001] = {};
    int arrb[100001] = {};

    scanf("%d %d %d ", &a, &b, &m);

    for (int i = 0; i < a; i++) {
        scanf("%d ", &arra[i]);
        amin = arra[0];
        if (min > arra[i])
            amin = arra[i];
    }
    for (int i = 0; i < b; i++) {
        scanf("%d ", &arrb[i]);
        bmin = arrb[0];
        if (min > arrb[i])
            bmin = arrb[i];
    }
    min = amin + bmin;

    int premin;
    for (int i = 0; i < m; i++) {
        x = y = c = 0;
        scanf("%d %d %d ", &x, &y, &c);
        premin = arra[x - 1] + arrb[y - 1] - c;
        if (premin < min)
            min = premin;
    }

    printf("%d\n", min);

    return 0;
}