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<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
#include<unordered_map>
#include<stack>
#include<string>
#include<iomanip>
#include<algorithm>
#include<functional>
#include<cstring>
#include<utility>
#include<math.h>
using namespace std;


/**** Type Define ****/

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> Q;

/**** Const List   ****/

const ll INF = 1LL << 60;
const ll mod = 1000000007;
const ll dx[4] = {1, 0, -1, 0};
const ll dy[4] = {0, -1, 0, 1};
const ll NCK_MAX = 510000;
/**** General Functions ****/
ll ketawa(ll n){
  ll a=0;
  while(n!=0){
    a+=n%10;
    n/=10;
  }
  return a;
}
ll RepeatSquaring(ll N, ll P, ll M){//pow->double
    if(P==0) return 1;
    if(P%2==0){
        ll t = RepeatSquaring(N, P/2, M);
        return (t% M)*(t % M)%M;
    }
    return (N * RepeatSquaring(N, P-1, M))%M;
}
ll RS(ll N, ll P){//modがだるいときにつかう
    if(P==0) return 1;
    if(P%2==0){
        ll t = RS(N, P/2);
        return t*t;
    }
    return (N * RS(N, P-1));
}
map< ll, int > prime_factor(ll n) {
  map< ll, int > ret;
  for(ll i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}
bool IsPrime(ll a){//order root a
  if(a==1)return false;
  for(int i=2;i*i<=a;i++){
    if(a%i==0&&a!=i){
      return false;
    }
  }
  return true;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll lcm(ll a, ll b){
    return a  / gcd(a, b)*b;
}

ll extgcd(ll a, ll b, ll& x, ll& y) {
  if (b == 0) {
    x = 1, y = 0; return a;
  }
  ll q = a/b, g = extgcd(b, a - q*b, x, y);
  ll z = x - q * y;
  x = y;
  y = z;
  return g;
}

ll invmod (ll a, ll m) { // a^-1 mod m
  ll x, y;
  extgcd(a, m, x, y);
  x %= m;
  if (x < 0) x += m;
  return x;
}

ll *fac, *finv, *inv;

void nCk_init(ll mod) {
  fac = new ll[NCK_MAX];
  finv = new ll[NCK_MAX];
  inv = new ll[NCK_MAX];
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (ll i = 2; i < NCK_MAX; i++) {
    fac[i] = fac[i-1] * i % mod;
    inv[i] = mod - inv[mod%i] * (mod / i) % mod;
    finv[i] = finv[i-1] * inv[i] % mod;
  }
}

ll nCk(ll n, ll k, ll mod) {
  if (fac == NULL) nCk_init(mod);
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n-k] % mod) % mod;
}

ll lmin(ll a, ll b) { return a > b ? b : a; };

ll lmax(ll a, ll b) { return a > b ? a : b; };

ll lsum(ll a, ll b) { return a + b; };

/**** Zip ****/

template <typename T>
class Zip {
  vector<T> d;
  bool flag;
public:
  Zip() {
    flag = false;
  }
  void add(T x) {
    d.push_back(x);
    flag = true;
  }
  ll getNum(T x) { // T need to have operator < !!
    if (flag) {
      sort(d.begin(), d.end());
      d.erase(unique(d.begin(), d.end()), d.end());
      flag = false;
    }
    return lower_bound(d.begin(), d.end(), x) - d.begin();
  }
  ll size() {
    if (flag) {
      sort(d.begin(), d.end());
      d.erase(unique(d.begin(), d.end()), d.end());
      flag = false;
    }
    return (ll)d.size();
  }
};

/**** Union Find ****/

class UnionFind {
  vector<ll> par, rank; // par > 0: number, par < 0: -par
public:
  void init(ll n) {
    par.resize(n, 1); rank.resize(n, 0);
  }
  ll getSize(ll x) {
    return par[find(x)];
  }
  ll find(ll x) {
    if (par[x] > 0) return x;
    return -(par[x] = -find(-par[x]));
  }
  void merge(ll x, ll y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (rank[x] < rank[y]) {
      par[y] += par[x];
      par[x] = -y;
    } else {
      par[x] += par[y];
      par[y] = -x;
      if (rank[x] == rank[y]) rank[x]++;
    }
  }
  bool isSame(ll x, ll y) {
    return find(x) == find(y);
  }
};

/**** Segment Tree ****/
class SegmentTree {
public:
  vector<pair<double, double>> node;//node[0]は使用しない
  ll n;//データの個数(nodeの最下層には何個並んでいるか)
  pair<double, double> initial_value;//初期値
public:
  void Init(ll n_,pair<double, double> initial_value_){
    n=1;
    while(n<n_)n*=2;
    node.resize(2*n);
    for(ll i=0;i<2*n;i++){
      node[i]=initial_value_;
    }
    initial_value=initial_value_;
  }
  void Update(ll k,pair<double, double> a){
    //node[k]をaにする
    //それに従って先祖も変わっていく
    k+=n;
    node[k]=a;
    while(k>1){
      k=k/2;
      node[k]=pair<double, double>(node[k*2].first*node[k*2+1].first,node[k*2].second*node[k*2+1].first+node[k*2+1].second);
    }
  }
  /*void Watch(){
    for(ll i=0;i<2*n;i++){
      cout<<node[i]<<endl;
    }
  }*/
  double Query(){//[a,b)
    return node[1].first+node[1].second;
  }
};


/**** LIS ****/
ll lis(ll* a, ll n, ll* dp) {
  fill(dp, dp + n, INF);//INFを代入
  for (ll i = 0; i < n; i++) *lower_bound(dp, dp + n, a[i]) = a[i];
  return (ll)(lower_bound(dp, dp + n, INF) - dp);
}


/**** main function ****/
ll a,b,m,n;
ll kosu[3];
ll ans=999999999999999;
string s;
vector<ll> graph[200001];
ll nibu[200001];
ll iti,ni;
ll herasu[3];
queue<ll> wan,tu,zero;
bool dfs(ll v,ll c){
  nibu[v]=c;
  for(ll i=0;i<graph[v].size();i++){
    if(nibu[graph[v][i]]==c) return false;//ありえん
    if((nibu[graph[v][i]]==0)&&!dfs(graph[v][i],-c)) return false;
  }
  return true;
}
int main(){
  cin>>n;
  kosu[0]=n/3;
  kosu[1]=(n+2)/3;
  kosu[2]=(n+1)/3;
  for(ll i=0;i<n-1;i++){
    ll aa,bb;
    cin>>aa>>bb;
    graph[bb].push_back(aa);
    graph[aa].push_back(bb);
  }

  dfs(1,1);
  for(ll i=1;i<=n;i++){
    if(nibu[i]==1)iti++;
    else ni++;
  }



  herasu[1]=iti-kosu[1];
  herasu[2]=ni-kosu[2];
  for(ll i=1;i<=n;i++){
    if(nibu[i]==1&&herasu[1]!=0){
      nibu[i]=0;
      herasu[1]--;
    }
    if(nibu[i]==-1&&herasu[2]!=0){
      nibu[i]=0;
      herasu[2]--;
    }
    if(i%3==1)wan.push(i);
    if(i%3==2)tu.push(i);
    if(i%3==0)zero.push(i);
  }
  /*if(herasu[1]!=0||herasu[2]!=0){
    cout<<-1<<endl;
    return 0;
  }*/
  for(ll i=1;i<=n;i++){
    if(nibu[i]==1){
      nibu[i]=wan.front();
      wan.pop();
    }else if(nibu[i]==-1){
      nibu[i]=tu.front();
      tu.pop();
    }else{
      nibu[i]=zero.front();
      zero.pop();
    }
  }
  for(ll i=1;i<=n;i++){
    if(nibu[i]==0){
      if(!wan.empty()){
        cout<<wan.front();
        wan.pop();
      }else if(!tu.empty()){
        cout<<tu.front();
        tu.pop();
      }
    }else{
      cout<<nibu[i];
    }

    if(i==n){
      cout<<endl;
    }else{
      cout<<" ";
    }
  }
}