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;

vector<int>g[200005];
int colr[200005];
int ans[200005];
void dfs(int cur,int par,int col){
   
    colr[cur] = col;  
    for(auto i: g[cur]){
      if( i != par) dfs(i,cur,3-col);
    }

}



int main(){

   int n;
   cin >> n;

   for(int i =0;i<n-1;i++){
        int u,v;cin >> u >> v;
        u--;
        v--;
        g[u].push_back(v);
        g[v].push_back(u);
   } 
   
   dfs(0,-1,1);
   int rc=0;
   int bc =0;
   for(int i =0;i<n;i++){
       if( colr[i] == 1) rc++;
       else bc++;
   }
  
  if( rc > bc){
     rc^=bc^=rc^=bc;
  }
  if( rc <= n/3){
    int k =1;
   
     for(int i =1;i<=n;i++){     
         if(colr[i-1] == 1 && 3*k<=n) {
           ans[i] = 3*k,k++;
         }
     }
     int ind = 0;
     for(int i =1;i<=n;i++){
       if( i>3*rc || i%3 && ans[ind]==0) ans[ind] =i,ind++;    
     }

   for(int i =1;i<=n;i++) cout << ans[i]<<" \n"[i==n]; 

  }
  else{
    int k =0;
    int l =0;
    for(int i =1;i<=n;i++){
       if(colr[i-1] == 1 && 3*k+1<=n) ans[i] = 3*k+1,k++;
       else if( colr[i-1] == 2 && 3*l+2<=n) ans[i] = 3*l +2,l++;
    }  
   // for(int i =1;i<=n;i++) cout << ans[i]<<" \n"[i==n];
    int m =1;

    for(int i =1;i<=n;i++){
      if(!ans[i]) ans[i] =3*m,m++;
    }
    
    for(int i =1;i<=n;i++) cout << ans[i]<<" \n"[i==n];

  }
 
}