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;
typedef long double ld;
typedef long long ll;
typedef pair<double, double> pdd;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef vector<pii> vii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
//const int mod = ;

vvi g;
vvi q(3);
vvi col(2);

void dfs(int v, int pr, int t) {
  col[t].push_back(v);
  for (int nv : g[v]) if (nv != pr) {
    dfs(nv, v, (t + 1) % 2);
  }
}

int main() {
  std::ios::sync_with_stdio(false); std::cin.tie(0);
  int n;
  cin >> n;
  g.resize(n);
  col.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a; --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= n; ++i) q[i % 3].push_back(i);
  dfs(0, -1, 0);
  vi res(n);
  if (col[0].size() > col[1].size()) swap(col[0], col[1]);
  cerr << col[0].size() << ' ' << col[1].size() << endl;
  if (col[0].size() <= q[0].size()) {
    for (int t = 0; t < 2; ++t) {
      for (int x : col[t]) {
        for (int r = 0; r < 3; ++r) if (!q[r].empty()) {
          res[x] = q[r].back();
          q[r].pop_back();
          break;
        }
      }
    }
  } else {
    for (int t = 0; t < 2; ++t) {
      for (int x : col[t]) {
        for (int r : {t + 1, 0}) if (!q[r].empty()) {
          res[x] = q[r].back();
          q[r].pop_back();
          break;
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) cout << res[i] << ' ';
  return 0;
}