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: Runtime Error)

#include <algorithm>
#include <bitset>
#include <cmath>
#include <complex>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <vector>

const int MOD = 1e9 + 7;
const int iINF = 2147483647 / 2;
const long long int llINF = 9223372036854775807 / 2;

using namespace std;
using ll = long long int;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;

typedef pair<ll, ll> pll;
bool paircomp(const pll &a, const pll &b) {
  if (a.first == b.first)
    return a.second < b.second;
  return a.first < b.first;
}

#define REP(i, n) for (ll i = 0; i < (n); i++)
#define RREP(i, n) for (ll i = (n)-1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define AUTO(i, m) for (auto &i : m)
#define ALL(a) (a).begin(), (a).end()
#define MAX(vec) *std::max_element(vec.begin(), vec.end())
#define MIN(vec) *std::min_element(vec.begin(), vec.end())
#define ARGMAX(vec)                                                            \
  std::distance(vec.begin(), std::max_element(vec.begin(), vec.end()))
#define ARGMIN(vec)                                                            \
  std::distance(vec.begin(), std::min_element(vec.begin(), vec.end()))
#define REV(T) greater<T>()
#define PQ(T) priority_queue<T, vector<T>, greater<T>>
#define VVL(a, b, c) vector<vector<ll>>(a, vector<ll>(b, c))
#define VV(T, a, b, c) vector<vector<T>>(a, vector<T>(b, c))
#define VVVL(a, b, c, d)                                                       \
  vector<vector<vector<ll>>>(a, vector<vector<ll>>(b, vector<ll>(c, d)))
#define VVV(T, a, b, c, d)                                                     \
  vector<vector<vector<T>>>(a, vector<vector<T>>(b, vector<T>(c, d)))
#define SP(a) setprecision(a)
#define SQRT(a) sqrt((long double)(a))
#define DPOW(a, b) pow((long double)(a), (long double)(b))
#define UNIQUE(vec)                                                            \
  do {                                                                         \
    sort(ALL((vec)));                                                          \
    (vec).erase(std::unique(ALL((vec))), (vec).end());                         \
  } while (0)

ll POW(ll n, ll m) {
  if (m == 0) {
    return 1;
  } else if (m % 2 == 0) {
    ll tmp = POW(n, m / 2);
    return (tmp * tmp);
  } else {
    return (n * POW(n, m - 1));
  }
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

void func(vvl &G, vl &C, ll now, ll par, ll s) {
  C[now] = s;
  AUTO(g, G[now]) {
    if (g != par) {
      func(G, C, g, now, (s + 1) % 2);
    }
  }
  return;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll N;
  cin >> N;
  vl A(N), B(N);
  vvl G(N);
  vl C(N, 0);
  vl ANS(N, -1);
  REP(i, N - 1) {
    cin >> A[i] >> B[i];
    G[A[i] - 1].push_back(B[i] - 1);
    G[B[i] - 1].push_back(A[i] - 1);
  }
  func(G, C, 0, -1, 0);
  ll z = 0;
  REP(i, N) {
    if (C[i] == 0)
      z++;
  }
  if (z <= N / 3) {
    vl po(N + 1, -1);
    ll count = 1;
    REP(i, N) {
      if (C[i] == 0) {
        while (count % 3 != 0)
          count++;
        po[count] = 1;
        ANS[i] = count;
        count++;
      }
    }
    count = 1;
    REP(i, N) {
      if (C[i] == 1) {
        while (po[count] == 1)
          count++;
        po[count] = 1;
        ANS[i] = count;
        count++;
      }
    }
  } else if (N - z <= N / 3) {
    // cout << z << endl;
    vl po(N + 1, -1);
    ll count = 1;
    REP(i, N) {
      if (C[i] == 1) {
        while (count % 3 != 0)
          count++;
        // cout << count << endl;
        po[count] = 1;
        ANS[i] = count;
        count++;
      }
    }
    count = 1;
    REP(i, N) {
      if (C[i] == 0) {
        while (po[count] == 1)
          count++;
        // cout << count << endl;
        po[count] = 1;
        ANS[i] = count;
        count++;
      }
    }
  } else {
    vl po(N + 1, -1);
    ll count = 1;
    REP(i, N) {
      if (C[i] == 0) {
        while (count % 3 == 1)
          count++;
        po[count] = 1;
        ANS[i] = count;
        count++;
      }
    }
    count = 1;
    REP(i, N) {
      if (C[i] == 1) {
        while (count % 3 == 2 || po[count] == 1)
          count++;
        po[count] = 1;
        ANS[i] = count;
        count++;
      }
    }
  }
  REP(i, N) cout << ANS[i] << endl;

  return 0;
}