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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define int long
using pi=pair<int,int>;
struct edge{
    int a, b;
    edge(){};
    edge(int x, int y)
        : a(x)
        , b(y){};
};
signed main() {
    int n;
    cin>>n;
    vector<edge> E(n - 1);
    vector<vector<int>> G(n);
    vector<int> dim(n, 0);
    for (int i = 0; i < n - 1; ++i) {
        cin >> E[i].a >> E[i].b;
        ++dim[--E[i].a], ++dim[--E[i].b];
        G[E[i].a].push_back(E[i].b);
        G[E[i].b].push_back(E[i].a);
    }
    int root = 0;
    vector<int> cand(n);
    for(int i=0;i<n;++i)cand[i]=i;
    sort(cand.begin(), cand.end(), [&](int a, int b) { return dim[a]%300 < dim[b]%300; });
    for (int i = 0; i < n; ++i) {
        root = cand[i];
        bool fail = false;
        queue<pi> que;
        vector<bool> vis(n, false);
        vector<int> res(n, -1);
        que.push(pi(root, 1));
        while (!que.empty()) {
            pi q = que.front();
            que.pop();
            res[q.first] = q.second;
            vis[q.first] = true;
            for (auto i : G[q.first]) {
                if (!vis[i]) que.push(pi(i, 3 - q.second));
            }
        }
        int u[3] = { 3, 1, 2 };
        for (int i = 0; i < n; ++i) {
            int k = res[i];
            if (u[k] <= n) {
                res[i] = u[k];
                u[k] += 3;
            } else
                res[i] = 0;
        }
        for (int i = 0; !fail && i < n; ++i) {
            if (!res[i]) {
                if (u[0] > n) {
                    fail = true;
                    break;
                }
                res[i] = u[0];
                u[0] += 3;
            }
        }
        if (!fail) {
            for (int i = 0; i < n; ++i) cout << (i ? " " : "") << res[i];
            cout << endl;
            return 0;
        }
    }
}