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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int n;
vector< vector< int > > edg;
vector< int > p;

void f(int x, int y){
    p[x] = y;
    for(int v : edg[x]){
        if(p[v] == -1){
            f(v, 3-y);
        }
    }
}


int main(void){
    cin >> n;
    edg.assign(n, vector<int>());
    for(int i=0;i<n-1;i++){
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        edg[a].push_back(b);
        edg[b].push_back(a);
    }

    for(int j=1;j<=2;j++){
        p.assign(n, -1);
        f(0, j);
        
        int x = 1, y = 2, z = 3;
        for(int i=0;i<n;i++){
            if(p[i] == 1){
                if(x <= n){
                    p[i] = x;
                    x += 3;
                }else{
                    p[i] = z;
                    z += 3;
                }
            }else{
                if(y <= n){
                    p[i] = y;
                    y += 3;
                }else{
                    p[i] = z;
                    z += 3;
                }
            }
        }

        if(x > n && y > n && z > n){
            break;
        }else if(j == 2){
            puts("-1");
            return 0;
        }
    }

    for(int i=0;i<n;i++){
        cout << p[i] << " ";
    }
    cout << endl;
}