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>
#define SIZE 200005
using namespace std;
typedef pair <int,int> P;
typedef long long int ll;

vector <int> vec[SIZE];
queue <int> vb,vw;
int ans[SIZE];

void dfs(int v=0,int p=-1,int d=0)
{
	if(d%2==0) vb.push(v);
	else vw.push(v);
	for(int i=0;i<vec[v].size();i++)
	{
		int to=vec[v][i];
		if(to!=p) dfs(to,v,d+1);
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n-1;i++)
	{
		int a,b;
		scanf("%d %d",&a,&b);a--,b--;
		vec[a].push_back(b);
		vec[b].push_back(a);
	}
	dfs();
	if(vb.size()<vw.size()) swap(vb,vw);
	for(int i=1;i<=n;i++)
	{
		if(i%3==1)
		{
			int v=vb.front();vb.pop();
			ans[v]=i;
		}
	}
	if(vb.size()<vw.size()) swap(vb,vw);
	for(int i=1;i<=n;i++)
	{
		if(i%3==2)
		{
			int v=vb.front();vb.pop();
			ans[v]=i;
		}
	}
	for(int i=1;i<=n;i++)
	{
		if(i%3==0)
		{
			int v;
			if(!vb.empty()) {v=vb.front();vb.pop();}
			else {v=vw.front();vw.pop();}
			ans[v]=i;
		}
	}
	for(int i=0;i<n;i++) printf("%d ",ans[i]);puts("");
	return 0;
}