Problem: 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.

Example

Input

5
1 2
1 3
3 4
3 5


Output

1 2 5 4 3

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<vector<int>> G;
int par[200010], dep[200010], ans[200010];

void dfs(int key){
	for(int i=0; i<G[key].size(); ++i){
		if(G[key][i]!=par[key]){
			par[G[key][i]]=key;
			dep[G[key][i]]=dep[key]+1;
			dfs(G[key][i]);
		}
	}
}

int main() {
	int N;
	cin >> N;
	G.resize(N+1);
	for(int i=1; i<N; ++i){
		int a, b;
		cin >> a >> b;
		G[a].emplace_back(b);
		G[b].emplace_back(a);
	}
	dep[1]=0;
	dfs(1);
	int od=0;
	for(int i=1; i<=N; ++i){
		od += dep[i]%2;
	}
	if(min(od, N-od)*3<=N){
		int xnow=3, ynow=1;
		for(int i=1; i<=N; ++i){
			if(dep[i]%2==(od*2<N ? 1 : 0)){
				ans[i]=xnow;
				xnow += 3;
			}else{
				ans[i]=ynow;
				if(ynow%3==0) ynow -= 3;
				else{
					ynow += ynow%3;
					if(ynow>N) ynow=(N/3)*3;
				}
			}
		}
	}else{
		int t=1, u=2, v=3;
		for(int i=1; i<=N; ++i){
			if(dep[i]%2==0){
				if(t<=N){
					ans[i]=t;
					t += 3;
				}else{
					ans[i]=v;
					v += 3;
				}
			}else{
				if(u<=N){
					ans[i]=u;
					u += 3;
				}else{
					ans[i]=v;
					v += 3;
				}
			}
		}
	}
	cout << ans[1];
	for(int i=2; i<=N; ++i) cout << ' ' << ans[i];
	cout << endl;
	return 0;
}