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;

enum color {red, blue};

vector<vector<int>> to(200000);
vector<int> v_red;
vector<int> v_blue;

void dfs(int v, color c, int prev)
{
  if (c == red) {
    v_red.push_back(v);
  } else {
    v_blue.push_back(v);
  }
  for (auto next : to.at(v)) {
    if (next == prev) continue;
    dfs(next, c == red ? blue : red, v);
  }
}

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

  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--; b--;
    to.at(a).push_back(b);
    to.at(b).push_back(a);
  }

  dfs(0, red, -1);

  vector<int> v = {1, 3, 2};
  if (v_red.size() > v_blue.size()) swap(v_red, v_blue);
  if (v_red.size() <= size_t(n / 3)) swap(v[0], v[1]);
  v_red.insert(end(v_red), begin(v_blue), end(v_blue));

  vector<int> v123;
  for (const auto& i : v) {
    for (int j = i; j <= n; j += 3) {
      v123.push_back(j);
    }
  }

  vector<int> ans(n);

  for (int i = 0; i < n; i++) {
    ans.at(v_red.at(i)) = v123.at(i);
  }

  for (int i = 0; i < n; i++) {
    cout << ans.at(i) << (i == n - 1 ? '\n' : ' ');
  }
}