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 col[200005],ans[200005],cnt[2];
vector<int> v[200005],pos[3];
void dfs(int node,int p)
{
	col[node]=(col[p]^1);
	cnt[col[node]]++;
	for (int u:v[node])
	{
		if (u!=p)
		dfs(u,node);
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=1;i<n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for (int i=1;i<=n;i++)
	pos[i%3].push_back(i);
	dfs(1,0);
	int s=0;
	if (cnt[1]<cnt[0])
	s=1;
	if (cnt[s]<=pos[0].size())
	{
		for (int i=1;i<=n;i++)
		{
			if (col[i]==s)
			{
				ans[i]=pos[0].back();
				pos[0].pop_back();
			}
		}
	}
	else
	{
		for (int i=1;i<=n;i++)
		{
			if (col[i]==s)
			{
				if (!pos[s+1].empty())
				{
					ans[i]=pos[s+1].back();
					pos[s+1].pop_back();
				}
				else
				{
					ans[i]=pos[0].back();
					pos[0].pop_back();
				}
			}
		}
	}
	for (int i=1;i<=n;i++)
	{
		if (!ans[i])
		{
			for (int j=0;j<3;j++)
			{
				if (!pos[j].empty())
				{
					ans[i]=pos[j].back();
					pos[j].pop_back();
					break;
				}
			}
		}
	}
	for (int i=1;i<=n;i++)
	printf("%d ",ans[i]);
}