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<bits/stdc++.h>
using namespace std;
int n,ans[200009];
vector<int>g[200009],mod[3],tp[2];
void dfs(int u,int p,bool c) {
	tp[c].push_back(u);
	for(int i=0;i<g[u].size();i++)
		if(g[u][i]!=p)
			dfs(g[u][i],u,!c);
}
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin>>n;
	for(int i=0;i<n-1;i++) {
		int a,b;
		cin>>a>>b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	for(int i=1;i<=n;i++) mod[i%3].push_back(i);
	dfs(1,0,0);
	if(tp[0].size()<tp[1].size()) swap(tp[0],tp[1]);
	if(tp[0].size()>=mod[1].size()+mod[2].size()) {
		while(mod[1].size()) {
			ans[tp[0].back()]=mod[1].back();
			tp[0].pop_back();
			mod[1].pop_back();
		}
		while(mod[2].size()) {
			ans[tp[0].back()]=mod[2].back();
			tp[0].pop_back();
			mod[2].pop_back();
		}
		for(int i=1;i<=n;i++)
			if(!ans[i]) {
				ans[i]=mod[0].back();
				mod[0].pop_back();
			}
	}
	else {
		while(mod[1].size()) {
			ans[tp[0].back()]=mod[1].back();
			tp[0].pop_back();
			mod[1].pop_back();
		}
		while(mod[2].size()) {
			ans[tp[1].back()]=mod[2].back();
			tp[1].pop_back();
			mod[2].pop_back();
		}
		for(int i=1;i<=n;i++)
			if(!ans[i]) {
				ans[i]=mod[0].back();
				mod[0].pop_back();
			}
	}
	for(int i=1;i<=n;i++) cout<<ans[i]<<' ';
	return 0;
}