Problem: 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.

Example

Input

5
1 2
1 3
3 4
3 5


Output

1 2 5 4 3

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

#define rep(i,n) for(int i = 0; i < (n); i++)
#define rrep(i,n) for(int i = (n)-1; i >= 0; i--)
#define rep1(i,n) for(int i = 1; i <= (n); i++)
#define rrep1(i,n) for(int i = (n); i > 0; i--)

#define ll long long
#define pi pair<int, int>
#define pll pair<ll, ll>

#define MOD 1000000007
#define INF 1000000000000000LL

using namespace std;

#define MAX_N 200000
vector<ll>g[MAX_N];

vector<int> group[2];
void dfs(ll pre, ll now, ll d){
  group[d].push_back(now);
  for(auto to: g[now]){
    if(pre!=to) dfs(now, to, d^1);
  }
}

int main(){
  ll n;cin>>n;
  rep(i, n-1){
    ll a,b;cin>>a>>b;
    a--;b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }

  dfs(-1, 0, 0);

  vector<int> num[3];
  rep1(i, n)num[i%3].push_back(i);

  vector<int> result(n);

  if(group[0].size() > group[1].size())swap(group[0], group[1]);

  if(group[0].size() <= n/3){
    for(auto tmp: group[0]){
      result[tmp] = num[0].back();
      num[0].pop_back();
    }
    for(auto tmp: group[1]){
      rep(i, 3){
        if(num[i].empty())continue;
        result[tmp] = num[i].back();
        num[i].pop_back();
        break;
      }
    }
    rep(i, n)cout<<result[i]<<' ';
    return 0;
  }

  rep(i, 2){
    for(auto tmp:group[i]){
      if( !num[i+1].empty() ){
        result[tmp] = num[i+1].back();
        num[i+1].pop_back();
      }else{
        result[tmp] = num[0].back();
        num[0].pop_back();
      }
    }
  }
  rep(i, n)cout<<result[i]<<' ';

  return 0;
}