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>
using namespace std;

#include <algorithm>
#include <iomanip>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
#include <random>
#include <numeric>

long long int n;
vector< vector< long long int > > v;

bool kansu( vector< long long int > & ans ) {

    std::random_device rnd;
    std::mt19937 mt(rnd());

    long long int cnt[3] = { ( n + 2 ) / 3, ( n + 1 ) / 3, n / 3 };
    long long int aa[3] = { 1, 2, 3 };

    vector< long long int > f(n);
    std::iota( f.begin(), f.end(), 0 );
    std::shuffle( f.begin(), f.end(), mt );

    while( cnt[0] + cnt[1] + cnt[2] > 0 ) {

        long long int i = f.back();
        f.pop_back();
        if ( ans[i] > 0 ) continue;

        if ( cnt[0] > cnt[1] ) {
            ans[i] = aa[0];
            cnt[0]--;
            aa[0] += 3;
        } else if ( cnt[0] < cnt[1] ) {
            ans[i] = aa[1];
            cnt[1]--;
            aa[1] += 3;
        } else if ( cnt[0] > 0 ) {
            ans[i] = aa[0];
            cnt[0]--;
            aa[0] += 3;
        } else {
            ans[i] = aa[2];
            cnt[2]--;
            aa[2] += 3;
            continue;
        }
        vector< long long int > ff(n);
        queue< long long int > q;
        q.push( i );
        ff[i] = 1;
        while( q.size() > 0 ) {
            long long int p = q.front();
            q.pop();
            long long int next = 3 - ( ans[p] % 3 );
            for ( long long int j = 0; j < v[p].size(); j++ ) {
                long long int k = v[p][j];
                if ( ans[k] > 0 ) continue;
                if ( ff[k] > 0 ) continue;
                ff[k] = ff[p] + 1;
                if ( ( ff[k] % 3 ) != 1 ) continue;
                if ( cnt[ next - 1 ] > 0 ) {
                    ans[k] = aa[ next - 1 ];
                    cnt[ next - 1 ]--;
                    aa[ next - 1 ] += 3;
                    q.push( k );
                    continue;
                }
                if ( cnt[2] > 0 ) {
                    ans[k] = aa[2];
                    cnt[2]--;
                    aa[2] += 3;
                    continue;
                }
                return false;
            }
        }
    }

    return true;

}

int main() {

    cin >> n;

    for ( long long int i = 0; i < n; i++ ) {
        vector< long long int > vv;
        v.push_back( vv );
    }

    for ( long long int i = 0; i < n - 1; i++ ) {
        long long int a, b;
        cin >> a >> b;
//a = 1; b = i + 2;
        a--;
        b--;
        v[a].push_back( b );
        v[b].push_back( a );
    }

    while( true ) {
        vector< long long int > ans(n);
        if ( !kansu( ans ) ) continue;

        for ( long long int i = 0; i < n; i++ ) {
            if ( i > 0 ) cout << " ";
            cout << ans[i];
        }
        cout << endl;
        break;
    }

    return 0;

}