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

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int INF = 1<<30;
const ll mod = 1000000007LL;

int main() {
  int N;
  cin>>N;
  vector<vector<int>> graph(N);
  for(int i = 0; i < N-1; i++){
    int x,y;
    cin>>x>>y;
    graph[x-1].push_back(y-1);
    graph[y-1].push_back(x-1);
  }
  queue<int> bfs;
  vector<int> col(N,-1);
  bfs.push(0);
  col[0]=0;
  vector<int> cnt(2,0);
  cnt[0]++;
  while(!bfs.empty()){
    int t = bfs.front();
    bfs.pop();
    for(auto nx:graph[t]){
      if(col[nx]!=-1) continue;
      col[nx]=1-col[t];
      bfs.push(nx);
      cnt[col[nx]]++;
    }
  }
  queue<int> a,b,c;
  for(int  i = 1; i <=N; i++){
    if(i%3==1)a.push(i);
    if(i%3==2)b.push(i);
    if(i%3==0)c.push(i);
  }
  vector<int> ans(N);
  if(cnt[0]>=(int)a.size()){
    for(int i = 0; i < N; i++){
      if(col[i]==0){
        if(a.empty()) {
          ans[i]=c.front();
          c.pop();
        }
        if(!a.empty()){
          ans[i]=a.front();
          a.pop();
        }
      }
      if(col[i]==1){
        if(b.empty()) {
          ans[i]=c.front();
          c.pop();
        }
        if(!b.empty()){
          ans[i]=b.front();
          b.pop();
        }
      }
    }
  }
  if(cnt[0]<(int)a.size()){
    for(int i = 0; i < N; i++){
      if(col[i]==0){
        ans[i]=c.front();
        c.pop();
      }
      if(col[i]==1){
        if(c.empty()){
          ans[i]=b.front();
          b.pop();
        }
        if(!c.empty()){
          ans[i]=c.front();
          c.pop();
        }
      }
    }
  }
  for(auto x:ans) cout<<x<<" ";
  cout<<endl;
}