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;

void solve(std::istream &in, std::ostream &out);

int main() {
#ifdef LOCAL
  freopen("../IO/c.in", "r", stdin);
//  freopen("../IO/c.out", "w", stdout);
#else
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
#endif
  solve(std::cin, std::cout);
  return 0;
}

int val[200000 + 555];
vector<int> g[200000 + 555];
vector<int> l, r;
void dfs(int v, int p, int d) {
  (d ? l : r).push_back(v);
  for (int u : g[v]) if (u != p) dfs(u, v, d ^ 1);
}

void solve(std::istream &in, std::ostream &out) {
  int n;
  in >> n;
  for (int i = 0, u, v; i < n - 1; ++i) {
    in >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0, 0);
  if ((int) l.size() > (int) r.size()) {
    swap(l, r);
  }
  int L = l.size();
  int st = 0;
  if (L <= n / 3) {
    int a = 3;
    for (int &i : l) {
      val[i] = a;
      st = a;
      a += 3;
    }
    a = 1;
    for (int &i : r) {
      if (a % 3 == 0) ++a;
      if (a > n) break;
      val[i] = a;
      ++a;
    }
  } else {
    int a = 1;
    for (int &i : l) {
      val[i] = a;
      if ((a += 3) > n) break;
    }
    a = 2;
    for (int &i : r) {
      val[i] = a;
      if ((a += 3) > n) break;
    }
  }
  for (int i = 1, a = st; i <= n; ++i) {
    cout << (val[i] ? val[i] : (a += 3)) << " \n"[i == n];
  }
}