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:
/*
luogu AT5759
*/
#include <bits/stdc++.h>
using namespace std;

int n;
int s[2],c[200005],num[200005],x1=1,x2=2,x3=3;
int ee,h[200005],nex[200005<<1],to[200005<<1];

void addedge(int x,int y)
{
	nex[++ee] = h[x];
	to[ee] = y;
	h[x] = ee;
}

void dfs(int x,int pre)
{
	c[x] = c[pre]^1;
	s[c[x]]++;
	for(int i=h[x];i;i=nex[i])
		if(to[i]!=pre)
			dfs(to[i],x);
}

int main()
{
	cin>>n;
	for(int i=1,x,y;i<n&&cin>>x>>y;i++)
		addedge(x,y), addedge(y,x);
	dfs(1,0);
	if(s[0]>n/3 && s[1]>n/3)
	{
		for(int i=1;i<=n;i++)
			if(c[i]==1)
			{
				if(x1<=n)
					num[i] = x1, x1 += 3;
				else
					num[i] = x3, x3 += 3;
			}
			else
			{
				if(x2<=n)
					num[i] = x2, x2 += 3;
				else
					num[i] = x3, x3 += 3;
			}
	}
	else if(s[0]<=n/3)
	{
		for(int i=1;i<=n;i++)
			if(c[i]==1)
			{
				if(x1<=n)
					num[i] = x1, x1 += 3;
				else if(x2<=n)
					num[i] = x2, x2 += 3;
				else if(x3<=n)
					num[i] = x3, x3 += 3;
			}
			else
				num[i] = x3, x3 += 3;
	}
	else if(s[1]<=n/3)
	{
		for(int i=1;i<=n;i++)
			if(c[i]==0)
			{
				if(x1<=n)
					num[i] = x1, x1 += 3;
				else if(x2<=n)
					num[i] = x2, x2 += 3;
				else if(x3<=n)
					num[i] = x3, x3 += 3;
			}
			else
				num[i] = x3, x3 += 3;
	}
	for(int i=1;i<=n;cout<<num[i++]<<' ');cout<<endl;
	
	return 0;
}