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;

int main() {
  ios::sync_with_stdio(false);

  int N;
  cin >> N;

  vector<vector<int>> G(N);
  for (int i = 0; i + 1 < N; ++i) {
    int A, B;
    cin >> A >> B;
    G[A - 1].push_back(B - 1);
    G[B - 1].push_back(A - 1);
  }

  function<pair<int, int>(int, int)> count = [&](int u, int f) {
    int c0 = 1;
    int c1 = 0;
    for (int v : G[u]) if (v != f) {
      int x, y;
      tie(x, y) = count(v, u);
      c0 += y;
      c1 += x;
    }
    return make_pair(c0, c1);
  };

  int c0, c1;
  tie(c0, c1) = count(0, -1);

  vector<int> res(N);
  int a = 1;
  int b = 2;
  int c = 3;
  function<void(int, int, int)> dfs = [&](int u, int f, int d) {
    if (min(c0, c1) > N / 3) {
      int &x = d ? a : b;
      if (x <= N) {
        res[u] = x;
        x += 3;
      } else {
        res[u] = c;
        c += 3;
      }
    } else if ((c0 <= N / 3 && !d) || (c1 <= N / 3 && d)) {
      res[u] = c;
      c += 3;
    }
    for (int v : G[u]) if (v != f) {
      dfs(v, u, d ^ 1);
    }
  };
  dfs(0, -1, 0);

  for (int &x : res) if (x == 0) {
    int &y = a <= N ? a : b <= N ? b : c;
    x = y;
    y += 3;
  }

  for (int i = 0; i < N; ++i) cout << res[i] << " \n"[i + 1 == N];

  return 0;
}