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 <iostream>
#include <vector>
#include <queue>
using namespace std;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N; cin >> N;
  vector<vector<int>> g(N+1);
  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);
  }

  queue<int> que;
  vector<int> dist(N, -1);
  vector<vector<int>> group(2);
  que.push(0);
  dist[0] = 0;  
  group[0].push_back(0);
  while (!que.empty()) {
    int v = que.front(); que.pop();
    for (int u : g[v]) {
      if (dist[u] < 0) {
        dist[u] = dist[v] + 1;
        group[dist[u] % 2].push_back(u);
        que.push(u);
      }
    }
  }

  vector<int> p(N), q = {1, 2, 3};
  if (group[1].size() <= N / 3) swap(group[0], group[1]);
  if (group[0].size() <= N / 3) {
    for (int v : group[0]) {
      p[v] = q[2]; q[2] += 3;
    }
    for (int v : group[1]) {
      if (q[2] <= N) { p[v] = q[2]; q[2] += 3; }
      else if (q[1] <= N) { p[v] = q[1]; q[1] += 3; }
      else { p[v] = q[0]; q[0] += 3; }
    }
  }
  else {
    for (int v : group[0]) {
      if (q[0] <= N) { p[v] = q[0]; q[0] += 3; }
      else { p[v] = q[2]; q[2] += 3; }
    }
    for (int v : group[1]) {
      if (q[1] <= N) { p[v] = q[1]; q[1] += 3; }
      else { p[v] = q[2]; q[2] += 3; }
    }
  }

  cout << p.front();
  for (int i = 1; i < N; i++) cout << " " << p[i];
  cout << "\n";
}