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 n;
int cnt[2] = {};
vector<vector<int>> edge, lst;
vector<int> res, memo;

void solve();
void dfs(int now, int dis);
void pattern3(bool ch);
void pattern12();

int main() {
  cin >> n;
  edge.resize(n);
  for(int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  solve();
  for(int i = 0; i < n; ++i) cout << res[i] << " ";
  cout << endl;
  return 0;
}

void solve() {
  res.resize(n);
  memo.assign(n, -1);
  dfs(0, 0);
  if(cnt[0] <= n / 3)
    pattern3(0);
  else if(cnt[1] <= n / 3)
    pattern3(1);
  else
    pattern12();
}

void pattern3(bool ch) {
  int nc[3] = {};
  for(int i = 0; i < n; ++i) {
    if(memo[i] % 2 == ch)
      res[i] = ++nc[2] * 3;
    else {
      res[i] = ++nc[0] * 3 - 2;
      if(res[i] > n) res[i] = ++nc[1] * 3 - 1;
      if(res[i] > n) res[i] = ++nc[2] * 3;
    }
  }
}

void pattern12() {
  int nc[3] = {};
  for(int i = 0; i < n; ++i) {
    res[i] = ++nc[memo[i] % 2] * 3 - 1 - memo[i] % 2;
    if(res[i] > n) res[i] = ++nc[2] * 3;
  }
}

void dfs(int now, int dis) {
  ++cnt[dis % 2];
  memo[now] = dis;
  for(auto p : edge[now])
    if(memo[p] == -1) dfs(p, dis + 1);
}