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)

#include <iostream>
#ifdef MBP
#include "stdc++.h"
#else
#include <bits/stdc++.h>
#endif

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define REP(i, n) for (long long int i = 0, i##_len = (n); i < i##_len; ++i)
#define REP1(i, n) for (long long int i = 1, i##_len = (n); i <= i##_len; ++i)
#define REP_R(i, n) for (long long int i = (n - 1); i >= 0; --i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((long long int)(x).size())
#define BIT(n) (1LL << (n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

//#define DEBUG

#ifdef DEBUG
#define OUT(x) std::cout << (x) << std::endl
#define OUT2(x, y) std::cout << (x) << "," << (y) << std::endl
#define OUT3(x, y, z) std::cout << (x) << "," << (y) << "," << (z) << std::endl
#define OUTSP(x) std::cout << (x) << " ";
#define OUTLF std::cout << std::endl
#define OUTPA(p) OUT2(p.first, p.second)
#else
#define OUT(x) NULL
#define OUT2(x, y) NULL
#define OUT3(x, y, z) NULL
#define OUTPA(p) NULL
#define OUTSP(p) NULL
#define OUTLF NULL
#endif

#define INF INT_MAX
#define INFLL LLONG_MAX
#define PB push_back
#define MOD 1000000007LL

#define LSB(x) __builtin_ctzll(x)
#define MSB(x) 64LL - __builtin_clzll(x) - 1LL
#define MSBINT(x) 32LL - __builtin_clz(x)
#define BITCNT(x) __builtin_popcountll(x)

using namespace std;
using uif = uint_fast64_t;
using ll = long long int;
using vec = vector<ll>;
using mat = vector<vector<ll>>;
using pa = pair<ll, ll>;
using tTree = __gnu_pbds::tree<ll, __gnu_pbds::null_type, less<ll>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;

template <class T>
inline void input(T &arr)
{
  ll n = arr.size();
  REP(i, n)
  cin >> arr[i];
}
template <class T>
inline void input2(T &arr1, T &arr2)
{
  ll n = arr1.size();
  REP(i, n)
  cin >> arr1[i] >> arr2[i];
}
template <class T>
inline void inputPA(T &arr)
{
  ll n = arr.size();
  REP(i, n)
  cin >> arr[i].first >> arr[i].second;
}
template <class T>
inline void outArr(T &arr)
{
  for (auto x : arr)
  {
    OUTSP(x);
  }
  OUTLF;
}

template <class T>
using PQ = priority_queue<T>;
template <class T>
using PQG = priority_queue<T, vector<T>, greater<T>>;

const ll dx[4] = {1, 0, -1, 0};
const ll dy[4] = {0, 1, 0, -1};
const ll dx8[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy8[8] = {0, 1, 1, 1, 0, -1, -1, -1};

template <class T>
bool chmax(T &, const T &);
template <class T>
bool chmin(T &, const T &);

ll qp(ll a, ll b);
ll qp(ll a, ll b, ll mo);

ll mod_pow(ll, ll);
ll mod_add(ll, ll);
ll mod_mul(ll, ll);
ll mod_div(ll, ll);
void make_fact(ll);
void make_invfact(ll);
void init_fact();

ll combination(ll, ll);

#define FACSIZE 200002LL

ll invfac[FACSIZE];
ll fac[FACSIZE];

struct edge
{
  ll to, cost, id;
};

class Graph
{
public:
  ll N;
  vector<vector<edge>> G;

  Graph(ll N) : N(N)
  {
    G = vector<vector<edge>>(N, vector<edge>(0));
  }

  void add_Directed_edge(ll from, ll to, ll cost = 1, ll id = 0)
  {
    G[from].push_back(edge({to, cost, id}));
  }

  void add_Undirected_edge(ll v1, ll v2, ll cost = 1, ll id = 0)
  {
    add_Directed_edge(v1, v2, cost, id);
    add_Directed_edge(v2, v1, cost, id);
  }

  mat waeshall_floyd()
  {
    mat d(N, vec(N, INF));
    REP(i, N)
    d[i][i] = 0;
    REP(i, N)
    for (edge e : G[i])
    {
      d[i][e.to] = min(d[i][e.to], e.cost);
    }
    REP(k, N)
    REP(i, N)
    REP(j, N)
    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);

    return d;
  }
  vec dikstra(ll x)
  {
    vec distance(N, INF);
    distance[x] = 0;
    PQG<pa> pque;
    pque.push(make_pair(0, x));

    while (!pque.empty())
    {

      auto pq = pque.top();
      pque.pop();
      for (auto e : G[pq.second])
      {
        ll d = distance[pq.second] + e.cost;
        if (d < distance[e.to])
        {
          distance[e.to] = d;
          pque.push(make_pair(d, e.to));
        }
      }
    }

    return distance;
  }
};

int main(void)
{
  ll n;
  cin >> n;
  vec A(n - 1);
  vec B(n - 1);
  Graph G(n);
  REP(i, n - 1)
  {
    cin >> A[i] >> B[i];
    ll a = A[i];
    ll b = B[i];
    G.add_Undirected_edge(a - 1, b - 1, 1, i);
  }
  auto d = G.dikstra(0);
  outArr(d);
  ll r = 0, b = 0;
  REP(i, n)
  {
    ll x = d[i] % 2;
    if (x)
      r++;
    else
      b++;
  }
  ll rv = 1;
  ll bv = 2;
  ll rest = 3;
  vec val(n, 0);
  set<ll> unse;
  REP(i, n)
  {
    unse.insert(i + 1);
  }

  if (b < n / 3)
  {
    OUT("D1");
    REP(i, n)
    {
      ll dis = d[i] % 2;
      if (dis == 0)
      {
        unse.erase(rv);
        val[i] = rv;
        rv += 3;
      }
      else
      {
        unse.erase(rest);
        val[i] = rest;
        rest += 3;
      }
    }
  }
  else if (r < n / 3)
  {
    OUT("D2");
    REP(i, n)
    {
      ll dis = d[i] % 2;
      if (dis != 0)
      {
        unse.erase(bv);
        val[i] = bv;
        bv += 3;
      }
      else
      {
        unse.erase(rest);
        val[i] = rest;
        rest += 3;
      }
    }
  }
  else
  {
    OUT("D3");
    REP(i, n)
    {
      ll dis = d[i] % 2;
      if (dis == 0)
      {
        unse.erase(rv);
        val[i] = rv;
        rv += 3;
      }
      else
      {
        unse.erase(bv);
        val[i] = bv;
        bv += 3;
      }
    }
  }

  outArr(val);
  outArr(unse);
  REP(i, n)
  {
    auto x = val[i];
    if (x > n)
    {
      auto us = *begin(unse);
      val[i] = us;
      unse.erase(us);
    }
  }
  //outArr(val);
  for (auto x : val)
  {
    cout << x << " ";
  }
  cout << endl;

  return 0;
}

template <class T>
bool chmax(T &a, const T &b)
{
  if (a < b)
  {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
  if (b < a)
  {
    a = b;
    return 1;
  }
  return 0;
}

ll qp(ll a, ll b)
{
  ll ans = 1LL;
  do
  {
    if (b & 1LL)
      ans = 1LL * mod_mul(ans, a) % MOD;
    a = 1LL * mod_mul(a, a) % MOD;
  } while (b >>= 1LL);
  return ans;
}
ll qp(ll a, ll b, ll mo)
{
  ll ans = 1LL;
  do
  {
    if (b & 1LL)
      ans = 1LL * (ans % mo) * (a % mo);
    a = 1LL * (a % mo) * (a % mo);
  } while (b >>= 1LL);
  return ans;
}

ll mod_pow(ll x, ll r)
{
  if (r == 0)
    return 1;
  else if (r == 1)
    return x % MOD;
  else if (r % 2 == 0)
  {
    ll t = mod_pow(x, r / 2) % MOD;
    return t * t % MOD;
  }
  else
  {
    ll t = mod_pow(x, r / 2) % MOD;
    ll k = t * t % MOD;
    return k * x % MOD;
  }
}
ll mod_add(ll a, ll b)
{
  return (a + b) % MOD;
}

ll mod_mul(ll a, ll b)
{
  return (a * b) % MOD;
}

ll mod_div(ll a, ll b)
{
  return mod_mul(a, mod_pow(b, MOD - 2));
}

void make_fact(ll n)
{
  fac[0] = 1;
  REP(i, n)
  {
    fac[i + 1] = mod_mul(fac[i], i + 1);
  }
}

void make_invfact(ll n)
{
  invfac[n] = mod_pow(fac[n], MOD - 2);
  REP_R(i, n)
  {
    invfac[i] = mod_mul(invfac[i + 1], i + 1);
  }
}

void init_fact()
{
  make_fact(FACSIZE - 1);
  make_invfact(FACSIZE - 1);
}

ll combination(ll n, ll c)
{

  return mod_mul(fac[n], mod_mul(invfac[c], invfac[n - c]));
}
///////////////fact
/*
  init_fact();
  cout << fac[5] << endl;
  cout << invfac[5] << endl;
  cout << combination(8, 4) << endl;
  cout << combination(31, 8) << endl;
  //*/
///////////////////