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;

const int N = 2e5 + 10;
vector<int> g[N], a[3], b[3];
int ans[N];

void dfs(int v = 0, int p = -1, int c = 1) {
  b[c].push_back(v);
  for (int to : g[v]) {
    if (to != p) {
      dfs(to, v, c ^ 3);
    }
  }
}

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

  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i) {
    a[i % 3].push_back(i);
  }
  dfs();
  for (int c = 1; c <= 2; ++c) {
    if (b[c].size() <= a[0].size()) {
      for (int v : b[c]) {
        ans[v] = a[0].back();
        a[0].pop_back();
      }
      for (int v : b[c ^ 3]) {
        int k = 0;
        while (a[k].empty()) {
          ++k;
        }
        ans[v] = a[k].back();
        a[k].pop_back();
      }
      for (int i = 0; i < n; ++i) {
        cout << ans[i] << ' ';
      }
      return 0;
    }
  }
  for (int c = 1; c <= 2; ++c) {
    for (int v : b[c]) {
      int k = c;
      if (a[k].empty()) {
        k = 0;
      }
      ans[v] = a[k].back();
      a[k].pop_back();
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }

  return 0;
}