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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <stdio.h>

int main(void) {
    int Num = 0;

    scanf("%d", &Num);

    int *a = NULL;
    int *b = NULL;

    a = (int*)malloc(sizeof(int)*(Num - 1));
    b = (int*)malloc(sizeof(int)*(Num - 1));

    for(int ii = 0; ii < (Num-1); ii++ ) {
        scanf("%d %d", &a[ii], &b[ii]);
    }

    int **tree = NULL;
    tree = (int**)malloc(sizeof(int*)*(Num-1));
    for(int ii = 0; ii < (Num - 1); ii++ ) {
        tree[ii] = (int*)malloc(sizeof(int)*(Num - 1));
    }

    //初期化
    for(int ii = 0; ii < (Num - 1); ii++ ) {
        for(int jj = 0; jj < (Num - 1); jj++ ) {
            tree[ii][jj] = -1;
        }
    }

    for(int ii = 0; ii < (Num - 1); ii++ ) {
        int reefNum = 0;

        if(a[ii] == (ii + 1) ) {
            tree[ii][reefNum] =b[ii];
            reefNum++; 
        }

        for(int jj = 0; jj < (Num-1); jj++) {
            if( b[jj] == (ii + 1) ) {
                tree[ii][reefNum] = a[jj];
                reefNum++;
            }
        }
    }

    for(int ii = 0; ii < (Num - 1); ii++ ) {
        for(int jj = 0; jj < (Num - 1); jj++ ) {
            if( tree[ii][jj] != -1 ) {
                int TreeNum1 = tree[ii][jj];

                for(int kk = 0; kk < (Num-1); kk++ ) {
                    if( tree[TreeNum1][kk] != -1 ) {
                        int TreeNum2 = tree[TreeNum1][kk];

                        for(int nn = 0; nn < (Num-1); nn++ ) {
                            int TreeNum3 = tree[TreeNum2][nn];
                            if( TreeNum3 != -1 ) {
                                std::cout << ii ;
                                std::cout << TreeNum3;
                            }

                        }

                    }
                }

            }
        }
    }

    if( b != NULL ) {
        free(b);
    }

    if( a!= NULL ) {
        free(a);
    } 

}