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;
using ll = long long;

#define REP(i,n) for(int i=0; i < (n); ++i)
#define REPR(i,n) for(int i=(n); i >= 0; --i)
#define FOR(i, m, n) for(int i = (m); i < (n); ++i)

// #define INF 1e9
vector<vector<int>> g;
vector<int> ans, depth;
void dfs(int v, int d){
  depth[v] = d;
  for(auto nv : g[v]){
    if(depth[nv] != -1) continue;
    dfs(nv, d+1);
  }
}
int main(){
  int N;
  cin >> N;
  g.resize(N);
  ans.resize(N), depth.resize(N, -1);
  REP(i, N-1){
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, 0);
  queue<int> mod0, mod1, mod2;
  FOR(i, 1, N+1){
    if(i % 3 == 0){
      mod0.push(i);
    }
    else if(i % 3 == 1){
      mod1.push(i);
    }
    else{
      mod2.push(i);
    }
  }
  REP(i, N){
    if(depth[i] % 2 == 0){
      if(!mod1.empty()){
        ans[i] = mod1.front(); mod1.pop();
      }
      else if(!mod0.empty()){
        ans[i] = mod0.front(); mod0.pop();
      }
      else{
        assert(0);
      }
    }
    else{
      if(!mod2.empty()){
        ans[i] = mod2.front(); mod2.pop();
      }
      else if(!mod0.empty()){
        ans[i] = mod0.front(); mod0.pop();
      }
      else{
        assert(0);
      }
    }
  }
  REP(i, N){
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}