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>
using namespace std;
#define MOD 1000000007

int main() {
  
  int N;
  cin >> N;

  vector<vector<int>> g(N,vector<int>(0));

  for(int i = 0;i < N-1;i++) {
    int a,b;
    cin >> a >> b;
    a--; b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }

  vector<int> color(N,0),seen(N,0);
  queue<int> Q;
  Q.push(0);
  seen[0] = 1;
  int w = 0,b = 0;
  w++;
  while(!Q.empty()) {
    int q = Q.front(); Q.pop();
    for(auto nv :g[q]) {
      if(seen[nv]) continue;
      Q.push(nv);
      seen[nv] = 1;
      if(color[q]) {
        color[nv] = 0;
        w++;
      }else {
        color[nv] = 1;
        b++;
      }
    }
  }

  int on = 1,tw = 2,th = 3;
  vector<int> ans(N,0);
  if(w > N/3 && b > N/3) {
    for(int i = 0;i < N;i++) {
      if(color[i]) {
        if(on <= N) {
          ans[i] = on;
          on += 3;
        }
      }else {
        if(tw <= N) {
          ans[i] = tw;
          tw += 3;
        }
      }
    }
  }else if(b <= N/3){
    for(int i = 0;i < N;i++) {
      if(color[i]) {
        ans[i] = th;
        th += 3;
      }
    }
  }else if(w <= N/3) {
    for(int i = 0;i < N;i++) {
      if(!color[i]) {
        ans[i] = th;
        th += 3;
      }
    }
  }

  set<int> S;
  for(int i = 1;i <= N;i++) S.insert(i);
  for(int i = 0;i < N;i++) S.erase(ans[i]);
  for(int i = 0;i < N;i++) {
    if(!ans[i]) {
      ans[i] = *S.begin();
      S.erase(S.begin());
    }
  }

  for(int i = 0;i < N;i++) cout << ans[i] << " ";
 
  cout << endl;
  

}