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;
using ll = long long;

const ll NMAX = 2e5 + 5;
ll N;
vector<vector<ll>> G(NMAX);
vector<ll> color(NMAX, 0);

void dfs(int c) {
  for (ll n: G[c]) {
    if (color[n]) continue;
    color[n] = 3 - color[c];
    dfs(n);
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  cin >> N;
  for (ll i = 0; i < N - 1; i++) {
    ll a, b; cin >> a >> b; a--; b--;
    G[a].emplace_back(b);
    G[b].emplace_back(a);
  }

  color[0] = 1;
  dfs(0);

  ll count1 = 0, count2 = 0;
  for (ll i = 0; i < N; i++) {
    if (color[i] == 1) count1++; else count2++;
  }

  ll x1 = 1, x2 = 2, x3 = 3;
  vector<ll> ans(N, 0);
  if (count1 <= N / 3 || count2 <= N / 3) {
    ll myColor = 1;
    if (count2 <= N / 3) myColor = 2;

    for (ll i = 0; i < N; i++) {
      if (color[i] == myColor) {
        ans[i] = x3;
        x3 += 3;
      } else if (x1 <= N) {
        ans[i] = x1;
        x1 += 3;
      } else if (x2 <= N) {
        ans[i] = x2;
        x2 += 3;
      } else {
        ans[i] = x3;
        x3 += 3;
      }
    }
  } else {
    for (ll i = 0; i < N; i++) {
      if (color[i] == 1 && x1 <= N) {
        ans[i] = x1;
        x1 += 3;
      } else if (color[i] == 2 && x2 <= N) {
        ans[i] = x2;
        x2 += 3;
      } else {
        ans[i] = x3;
        x3 += 3;
      }
    }
  }
  cout << ans[0]; for (int i = 1; i < ans.size(); i++) { cout << ' ' << ans[i]; }

  return 0;
}