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;
vector<int>v[3],g[2],l[200003]; 
int n,ans[200003];bool used[200003];
void dfs(int x,bool a,int p){
	g[a].push_back(x);
	for(int i=0;i<l[x].size();i++)
		if(l[x][i]!=p)
			dfs(l[x][i],a^1,x);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<n;i++){int x,y;scanf("%d%d",&x,&y);x--;y--;l[x].push_back(y);l[y].push_back(x);}
	dfs(0,0,-1);
	for(int i=1;i<=n;i++)v[i%3].push_back(i);
	if(g[1].size()<g[0].size())swap(g[0],g[1]);
	if(g[0].size()<=v[0].size())
	{
		for(int i=0;i<g[0].size();i++)
			ans[g[0][i]]=v[0][i],
			used[v[0][i]]=1;
		for(int i=1;i<=n;i++)
			if(!used[i])
				ans[g[1].back()]=i,
				g[1].pop_back();
	}else{
		for(int i=0;i<v[1].size();i++)
			ans[g[0].back()]=v[1][i],
			g[0].pop_back();
		for(int i=0;i<v[2].size();i++)
			ans[g[1].back()]=v[2][i],
			g[1].pop_back();
		for(int i=0;i<g[0].size();i++)
			ans[g[0][i]]=v[0].back(),
			v[0].pop_back();
		for(int i=0;i<g[1].size();i++)
			ans[g[1][i]]=v[0].back(),
			v[0].pop_back();
	}
	for(int i=0;i<n;i++)printf("%d ",ans[i]);
}