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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef pair<ll,ll> pi;
typedef vector<pi> vpi;
typedef long double ld;
#define pb emplace_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define ALL(x) x.begin(), x.end()
#define SZ(x) (ll)x.size()
#define f first
#define s second
#define MAXN 400010
const ll MOD = 998244353;
const ll INF = 1e15;

ll N;
vi V[MAXN];
ll a,b;
ll dst[MAXN];
vi O;
vi E;
ll out[MAXN];

void dfs(int x, int p){
	for (auto v:V[x])if(v!=p){
		dst[v] = dst[x]+1;
		dfs(v,x);
	}
}

int main(){
	cin>>N;
	for (int i=0;i<=N;++i){
		cin>>a>>b;
		V[a].pb(b);
		V[b].pb(a);
	}
	dfs(1,-1);
	int t = 1;
	for (int i=1;i<=N;++i){
		if (dst[i]%2==1)O.pb(i);
		else E.pb(i);
	}
	for (auto i : O){
		if (t>N)break;
		out[i] = t;
		t+=3;
	}
	t = 2;
	for (auto i : E){
		if (t>N)break;
		out[i] = t;
		t+=3;
	}
	t = 3;
	for (int i=1;i<=N;++i)if(out[i] == 0){
		if (t>N){
			cout<<-1;
			return 0;
		}
		out[i] = t;
		t += 3;
	}
	for (int i=1;i<=N;++i)cout<<out[i]<<' ';
}