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:
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
constexpr int N = 202020;
vector<int> g[N];
int col[N];
vector<int> v[2];
void dfs (int cur, int par) {
  for (int child : g[cur]) if (child != par) {
    col[child] = col[cur] ^ 1;
    dfs (child, cur);
  }
}
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  fill(col, col + n, -1);
  col[0] = 0;
  dfs (0, -1);
  for (int i = 0; i < n; i++) v[col[i]].push_back(i);
  if ((int) v[0].size() < (int) v[1].size()) v[0].swap(v[1]);
  int cnt[] = {n / 3 + ((n % 3) > 0), n / 3 + ((n % 3) > 1), n / 3};
  vector<int> ans(n);
  for (int i = 0; i < n && v[0].size(); i += 3, v[0].pop_back()) {
    ans[v[0].back()] = i;
  }
  if ((int) v[0].size() >= cnt[1]) {
    for (int i = 1; i < n && v[0].size(); i += 3, v[0].pop_back()) {
      ans[v[0].back()] = i;
    }
  } else {
    for (int i = 1; i < n && v[1].size(); i += 3, v[1].pop_back()) {
      ans[v[1].back()] = i;
    }
  }
  for (int i = 2; i < n; i += 3) {
    if (v[1].size()) ans[v[1].back()] = i, v[1].pop_back();
    else ans[v[0].back()] = i, v[0].pop_back();
  }
  for (int i = 0; i < n; i++) {
    if (i > 0) cout << " ";
    cout << ans[i] + 1;
  }
  cout << endl;
  return 0;
}