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 <bits/stdc++.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); ++i)
#define repr(i, n) for(int i = (n-1); i >= 0; --i)

typedef long long ll;
typedef pair<int, int> Pi;

const int MAX = 2 * 1e5 + 5;
const int INF = 1001001001;
const int MOD = 1000000007;

vector<int> G[MAX];
vector<int> D;

void dfs(int n, int c=0) {
    if (D[n] < 0) D[n] = c;
    else return;
    for (int &i : G[n]) {
        dfs(i, 1-c);
    }
}

int main(){
    int N;
    cin >> N;

    rep(i, N) {
        int a, b;
        cin >> a >> b;
        a--, b--;
        G[a].push_back(b);
        G[b].push_back(a);
    }

    D = vector<int>(N, -1);
    dfs(0);

    int num[3] = {3, 1, 2};
    vector<int> ans(N);
    rep(i, N) {
        if (D[i] == 0) {
            if (num[1] <= N) {
                ans[i] = num[1];
                num[1] += 3;
            } else {
                ans[i] = num[0];
                num[0] += 3;
            }
        } else {
            if (num[2] <= N) {
                ans[i] = num[2];
                num[2] += 3;
            } else {
                ans[i] = num[0];
                num[0] += 3;
            }
        }
    }

    for (int &i : ans) cout << i << " ";
    cout << endl;
}