Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

5
1 2
1 3
3 4
3 5

Sample Output 1

1 2 5 4 3

The distance between two vertices is 3 for the two pairs (2, 4) and (2, 5).

p_2 + p_4 = 6

p_2\times p_5 = 6

Thus, this permutation satisfies the condition.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define LOCAL
#define _USE_MATH_DEFINES
#include <array>
#include <cassert>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <iomanip>
#include <string>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <complex>
#include <cmath>
#include <numeric>
#include <bitset>
#include <functional>
#include <random>
#include <ctime>

using namespace std;

template <typename A, typename B>
ostream& operator <<(ostream& out, const pair<A, B>& a) {
  out << "(" << a.first << "," << a.second << ")";
  return out;
}
template <typename T, size_t N>
ostream& operator <<(ostream& out, const array<T, N>& a) {
  out << "["; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
  return out;
}
template <typename T>
ostream& operator <<(ostream& out, const vector<T>& a) {
  out << "["; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
  return out;
}
template <typename T, class Cmp>
ostream& operator <<(ostream& out, const set<T, Cmp>& a) {
  out << "{"; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "}";
  return out;
}
#ifdef LOCAL
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...) 42
#endif
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}

typedef long long int64;
typedef pair<int, int> ii;
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x; }

struct fast_ios {
  fast_ios() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
  };
} fast_ios_;

const int N = 2e5 + 10;
int c[N];
vector<int> a[N];
vector<int> A[3];

void DFS(int u, int parent) {
  for (auto& v : a[u]) {
    if (v == parent) continue;
    c[v] = 3 - c[u];
    DFS(v, u);
  }
}

int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x; --y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  c[0] = 1;
  DFS(0, -1);
  for (int i = 0; i < n; ++i) {
    A[c[i]].push_back(i);
  }
  vector<int> cnt(3);
  vector<int> ret(n);
  for (int i = 1; i <= n; ++i) cnt[i % 3]++;
  if (A[1].size() <= cnt[0]) {
    int x = 3;
    for (auto& u : A[1]) {
      ret[u] = x;
      x += 3;
    }
    x = 1;
    for (int i = 0; i < n; ++i) {
      if (ret[i]) continue;
      if (x % 3 == 0) x += 1;
      ret[i] = x++;
    }
  } else if (A[2].size() <= cnt[0]) {
    int x = 3;
    for (auto& u : A[2]) {
      ret[u] = x;
      x += 3;
    }
    x = 1;
    for (int i = 0; i < n; ++i) {
      if (ret[i]) continue;
      if (x % 3 == 0) x += 1;
      ret[i] = x++;
    }
  } else {
    int x = 3;
    for (int i = 0; i < cnt[1]; ++i) {
      ret[A[1][i]] = 3 * i + 1;
    }
    for (int i = cnt[1]; i < A[1].size(); ++i) {
      ret[A[1][i]] = x;
      x += 3;
    }
    for (int i = 0; i < cnt[2]; ++i) {
      ret[A[2][i]] = 3 * i + 2;
    }
    for (int i = cnt[2]; i < A[2].size(); ++i) {
      ret[A[2][i]] = x;
      x += 3;
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%d%c", ret[i], " \n"[i + 1 == n]);
  }
  return 0;
}