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 maxn = 2e5 + 10;

int N;
vector<int> adj[maxn];
vector<int> color[2];
vector<int> result;

void dfs(int u, int p, int c) {
  color[c].push_back(u);
  for (int v : adj[u]) if (v != p) {
    dfs(v, u, c ^ 1);
  }
}

void calc(int &id, int r) {
  if (color[id].size() < color[id ^ 1].size()) id ^= 1;
  for (int i = 1; i <= N; ++i) if (i % 3 == r) {
    result[color[id].back()] = i;
    color[id].pop_back();
  }
}

int main() {
  ios_base::sync_with_stdio(0); cin.tie(NULL);
  cin >> N;
  for (int i = 1; i < N; ++i) {
    int u, v; cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0, 0);
  result.assign(N + 1, 0);
  int id = 0;
  calc(id, 1);
  calc(id, 2);
  for (int i = 1; i <= N; ++i) if (i % 3 == 0) {
    int v;
    if (!color[id].empty()) {
      v = color[id].back(); color[id].pop_back();
    } else {
      v = color[id ^ 1].back(); color[id ^ 1].pop_back();
    }
    result[v] = i;  
  }
  for (int i = 1; i <= N; ++i) cout << result[i] << " ";
  return 0;
}