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 pb push_back

const int N=200005;

int n,a[N],ans[N];
std::vector<int> e[N],vec[2];

void dfs(int x,int fa,int d)
{
	vec[d].pb(x);
	for (int to:e[x])
		if (to!=fa) dfs(to,x,d^1);
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		e[x].pb(y);e[y].pb(x);
	}
	dfs(1,0,0);
	int k1=n/3,k2=n/3;
	if (n%3==1) k1++;
	else if (n%3==2) k1++,k2++;
	if (vec[0].size()>vec[1].size()) std::swap(vec[0],vec[1]);
	if (k1<=vec[0].size()&&k2<=vec[1].size())
	{
		for (int i=0;i<k1;i++) ans[vec[0][i]]=1;
		for (int i=0;i<k2;i++) ans[vec[1][i]]=2;
	}
	else
	{
		for (int i=0;i<k1;i++) ans[vec[1][i]]=1;
		for (int i=k1;i<k1+k2;i++) ans[vec[1][i]]=2;
	}
	int now0=3,now1=1,now2=2;
	for (int i=1;i<=n;i++)
		if (!ans[i]) a[i]=now0,now0+=3;
		else if (ans[i]==1) a[i]=now1,now1+=3;
		else a[i]=now2,now2+=3;
	for (int i=1;i<=n;i++) printf("%d ",a[i]);
	return 0;
}