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<cstdio>
#include<vector>
using namespace std;
const int N=200005;
int n;
vector<int>G[N];
int dep[N];
vector<int>pos[2];
void dfs(int u,int father)
{
	dep[u]=dep[father]+1;
	pos[dep[u]&1].push_back(u);
	for(int v:G[u])
	{
		if(v==father) continue;
		dfs(v,u);
	}
	return;
}
vector<int>col[3];
int ans[N];
int main()
{
	scanf("%d",&n);
	int num[3]={n/3,(n+2)/3,(n+1)/3};
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		G[x].push_back(y);
		G[y].push_back(x);
	}
	dfs(1,0);
	int x=pos[0].size(),y=pos[1].size();
	if(x<=num[0])
	{
		for(int u:pos[0])
			col[0].push_back(u),num[0]--;
		for(int u:pos[1])
		{
			if(num[0]>0) col[0].push_back(u),num[0]--;
			else if(num[1]>0) col[1].push_back(u),num[1]--;
			else if(num[2]>0) col[2].push_back(u),num[2]--;
		}
	}
	else if(y<=num[0])
	{
		for(int u:pos[1])
			col[0].push_back(u),num[0]--;
		for(int u:pos[0])
		{
			if(num[0]>0) col[0].push_back(u),num[0]--;
			else if(num[1]>0) col[1].push_back(u),num[1]--;
			else if(num[2]>0) col[2].push_back(u),num[2]--;
		}
	}
	else
	{
		for(int u:pos[0])
		{
			if(num[1]>0) col[1].push_back(u),num[1]--;
			else if(num[0]>0) col[0].push_back(u),num[0]--;
		}
		for(int u:pos[1])
		{
			if(num[2]>0) col[2].push_back(u),num[2]--;
			else if(num[0]>0) col[0].push_back(u),num[0]--;
		}
	}
	int now=1;
	for(int u:col[1])
		ans[u]=now,now+=3;
	now=2;
	for(int u:col[2])
		ans[u]=now,now+=3;
	now=3;
	for(int u:col[0])
		ans[u]=now,now+=3;
	for(int i=1;i<=n;i++)
		printf("%d ",ans[i]);
	return 0;
}