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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <iomanip>
#include <vector>
#include <utility>
#include <map>
#include <algorithm>
#include <queue>
#include <cmath>
#include <numeric>
#include <set>

using namespace std;
struct aaa{aaa(){cin.tie(nullptr); ios::sync_with_stdio(false); cout<<fixed<<setprecision(20);};}aaa;
template <class T>ostream &operator<<(ostream &o,const vector<T>&v){o<<"{";for(int i=0;i<(int)v.size();i++)o<<(i>0?", ":"")<<v[i];o<<"}";return o;}
#define debug(v) {cerr<<"\033[1;36m[debug]\033[m "<<#v<<" : "<<v<<endl;}

using int64 = long long;
using V = vector<int>;

void dfs(const int& u, const bool& b, const int& p, V& bipartite, const vector<V>& g, vector<V>& bi) {
    if (bipartite[u] > 0) return;
    bipartite[u] = b;
    bi[b].push_back(u);

    for (auto v : g[u]) {
        if (v == p) continue;
        dfs(v, !b, u, bipartite, g, bi);
    }
}

int main() {
    int n;
    cin >> n;
    vector<V> g(n+1);

    for (int i=1; i<n; i++) {
        int a,b;
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
    }


    V bipartite(n+1, -1);
    vector<V> bi(2);

    dfs(1, false, -1, bipartite, g, bi);
    vector<vector<int>> m(3);
    for (int i=1; i<=n; i++) {
        m[i%3].push_back(i);
    }

    if (bi[0].size() > bi[1].size()) swap(bi[0], bi[1]);
    
    vector<int> ans(n+1);
    for (int i=1; i<=2; i++) {
        for (auto b : bi[i-1]) {
            if (m[i].size() > 0) {
                ans[b] = m[i].back();
                m[i].pop_back();
            } else {
                ans[b] = m[0].back();
                m[0].pop_back();
            }
        }
    }

    for (int i=1; i<=n; i++)
        cout << ans[i] << (i!=n? " " : "\n");

}