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: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <iomanip>
#include <math.h>
#include <fstream>

using namespace std;

// long long mod = 998244353;
long long mod = 1000000007;
vector<long long> fact(1000001);
vector<long long> invFact(1000001);

vector<int> setSize(100000, 1);

long long gcd(long long a, long long b)
{
    if (b == 0)
    {
        return a;
    }
    return gcd(b, a % b);
}

long long binpow(long long x, long long k)
{
    if (k == 0)
    {
        return 1LL;
    }
    long long temp = binpow(x, k / 2);
    if (k % 2 == 1)
    {
        return (((temp * temp) % mod) * x) % mod;
    }
    else
    {
        return (temp * temp) % mod;
    }
}

long long modular_inverse(long long x)
{
    return binpow(x, mod - 2);
}

long long binomial(long long n, long long k)
{
    long long rez = 1;
    for (long long i = n - k + 1; i <= n; i++)
    {
        rez *= i;
        rez %= mod;
    }
    long long temp = 1;
    for (long long i = 1; i <= k; i++)
    {
        temp *= i;
        temp %= mod;
    }
    rez *= modular_inverse(temp);
    rez %= mod;
    return rez;
}

int fi(int a, vector<int> &p)
{
    if (p[a] != p[p[a]])
    {
        p[a] = fi(p[a], p);
    }
    return p[a];
}

void uni(int a, int b, vector<int> &p)
{
    int x = fi(a, p), y = fi(b, p);
    if (x != y)
    {
        p[x] = y;
        setSize[y] += setSize[x];
    }
}

long long euclidianDistanceSquared(pair<int, int> a, pair<int, int> b)
{
    return (a.first - b.first) * (a.first - b.first) + (a.second - b.second) * (a.second - b.second);
}

double euclidianDistanceSquared(pair<double, double> a, pair<double, double> b)
{
    return (a.first - b.first) * (a.first - b.first) + (a.second - b.second) * (a.second - b.second);
}

double euclidianDistance(pair<double, double> a, pair<double, double> b)
{
    return sqrt(euclidianDistanceSquared(a, b));
}

bool circleIntersections(pair<double, double> pos1, double r1, pair<double, double> pos2, double r2, pair<double, double> &int1, pair<double, double> &int2)
{
    double d = euclidianDistance(pos1, pos2);
    if (r1 + r2 < d || d < fabs(r1 - r2))
    {
        return false;
    }
    double a = (r1 * r1 - r2 * r2 + d * d) / (2 * d);
    double h = sqrt(r1 * r1 - a * a);
    pair<double, double> pos3 = { pos1.first + a * (pos2.first - pos1.first) / d, pos1.second + a * (pos2.second - pos1.second) / d };
    int1.first = pos3.first + h * (pos2.second - pos1.second) / d;
    int1.second = pos3.second - h * (pos2.first - pos1.first) / d;
    int2.first = pos3.first - h * (pos2.second - pos1.second) / d;
    int2.second = pos3.second + h * (pos2.first - pos1.first) / d;
    return true;
}

class SegmentTree
{
public:
    class Node
    {
    public:
        vector<int> value;
        Node *l, *r;
        int left, right;

        Node(int x, int y)
        {
            value = vector<int>(26, 0);
            l = nullptr;
            r = nullptr;
            left = x;
            right = y;
        }

        void update(int p, char c)
        {
            if (left > p || p > right)
            {
                return;
            }
            if (left == p && p == right)
            {
                for (int i = 0; i < 26; i++)
                {
                    value[i] = 0;
                }
                value[c - 'a'] = 1;
                return;
            }
            if (l == nullptr)
            {
                int mid = (left + right) / 2;
                l = new Node(left, mid);
                r = new Node(mid + 1, right);
            }
            l->update(p, c);
            r->update(p, c);
            for (int i = 0; i < 26; i++)
            {
                value[i] = l->value[i] + r->value[i];
            }
        }

        int query(int x, int y)
        {
            if (left > y || x > right)
            {
                return 0;
            }
            if (x <= left  && right <= y)
            {
                int rez = 0;
                for (int i = 0; i < 26; i++)
                {
                    if (value[i] != 0)
                    {
                        rez |= (1 << i);
                    }
                }
                return rez;
            }
            if (l == nullptr)
            {
                return 0;
            }
            return l->query(x, y) | r->query(x, y);
        }
    };

    Node *root;
    SegmentTree(int n)
    {
        root = new Node(0, n - 1);
    }

    void update(int p, char c)
    {
        root->update(p, c);
    }

    int query(int x, int y)
    {
        return root->query(x, y);
    }
};

void dfs(int curr, int prev, vector<vector<int>> &v, vector<int> &colors)
{
    if (prev == -1)
    {
        colors[curr] = 1;
    }
    else
    {
        colors[curr] = (colors[prev] + 1) % 2;
    }
    for (int i = 0; i < v[curr].size(); i++)
    {
        if (v[curr][i] == prev)
        {
            continue;
        }
        dfs(v[curr][i], curr, v, colors);
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(12);
    // ifstream cin("in.txt");
    // ofstream cout("out.txt");
    
    
    int n;
    cin >> n;
    vector<vector<int>> v(n);
    vector<int> colors(n, -1);
    for (int i = 0; i < n - 1; i++)
    {
        int a, b;
        cin >> a >> b;
        v[a - 1].push_back(b - 1);
        v[b - 1].push_back(a - 1);
    }
    dfs(0, -1, v, colors);
    vector<int> res(n, 0);
    int reds = 0, blues = 0;
    for (int i = 0; i < n; i++)
    {
        if (colors[i] == 1)
        {
            reds++;
        }
        else
        {
            blues++;
        }
    }
    if (blues < reds)
    {
        for (int i = 0; i < n; i++)
        {
            (colors[i] += 1) %= 2;
        }
        swap(blues, reds);
    }
    if (reds <= n / 3)
    {
        int x = 1, y = 3;
        for (int i = 0; i < n; i++)
        {
            if (colors[i] == 1)
            {
                cout << y << " ";
                y += 3;
            }
            else
            {
                if (x > n)
                {
                    cout << y << " ";
                    y += 3;
                    continue;
                }
                cout << x << " ";
                x++;
                if (x % 3 == 0)
                {
                    x++;
                }
            }
        }
    }
    else
    {
        int a = 1, b = 2, c = 3;
        for (int i = 0; i < n; i++)
        {
            if (colors[i] == 1)
            {
                if (a <= n)
                {
                    res[i] = a;
                    a += 3;
                }
                else
                {
                    res[i] = c;
                    c += 3;
                }
            }
            else
            {
                if (b <= n)
                {
                    res[i] = b;
                    b += 3;
                }
                else
                {
                    res[i] = c;
                    c += 3;
                }
            }
            cout << res[i] << " ";
        }
    }
    
    




    return 0;
}