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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define sc(x) scanf("%lld",&(x));
#define pb push_back
#define fi first
#define se second
#define P pair<int,int>
#define endl '\n'
int n;
vector<int>G[200005];

int a=0,b=0;
int d[200005];
void dfs(int x,int y,int fa)
{
    if(y==1)
    {
        d[x]=1,a++;
    }
    else d[x]=0,b++;
    for(int i=0; i<G[x].size(); i++)
    {
        if(G[x][i]!=fa)
            dfs(G[x][i],y^1,x);
    }
}
int ans[200005];
signed main()
{
    sc(n)
    int x,y;
    for(int i=0; i<n-1; i++)
    {
        sc(x)sc(y)
        G[x].pb(y);
        G[y].pb(x);
        //2*n-2
    }

    dfs(1,1,-1);
    if(b <= n/3){
        a = b;
        for(int i=1;i<=n;i++)d[i]^=1;
    }
    if(a <= n/3)
    {
        int o = 0, p = 0, q = 0;
        for(int i=1; i<=n; i++)
        {
            if(d[i]==0)
            {
                ans[i]=o*3+3;
                o++;
            }
            else
            {
                if(p*3+1<=n)
                {
                    ans[i]=p*3+1;
                    p++;
                }
                else if(q*3+2<=n)
                {
                    ans[i]=q*3+2;
                    q++;
                }
                else if(o*3+3<=n)
                {
                    ans[i]=o*3+3;
                    o++;
                }
            }
        }
        for(int i=1;i<=n;i++){
            cout<<ans[i]<<' ';
        }
        return 0;
    }

    int k=0,o=0,q=0;
    for(int i=1; i<=n; i++)
    {
        if(d[i]==1)
        {
            if(k*3+1>n)
            {
                if(q*3+3>n)
                {
                    continue;

                }
                ans[i]=q*3+3;
                q++;
                continue;
            }
            ans[i]=k*3+1;

            k++;
        }
        else
        {
            if(o*3+2>n)
            {
                if(q*3+3>n)
                {
                    continue;
                }
                ans[i]=q*3+3;
                q++;
                continue;
            }
            ans[i]=o*3+2;
            o++;
        }
    }
    for(int i=1; i<=n; i++)
    {
        if(ans[i]==0&&k*3+1<=n)
        {
            ans[i]=k*3+1;
            k++;
        }
        if(ans[i]==0&&o*3+2<=n)
        {
            ans[i]=o*3+2;
            o++;
        }
    }
    for(int i=1; i<=n; i++)cout<<ans[i]<<' ';

}