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:
//love yjl forever
#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
const int INF=0x3f3f3f3f,N=2e6+5,M=2e5+5;
int n;
int head[N],to[N*4],Next[N*4],cnt;
void add(int x,int y)
{
	Next[++cnt]=head[x];head[x]=cnt;to[cnt]=y;
}
int dep[M],ans[M];
vector<int> v[3];
void dfs(int x,int fa)
{
	dep[x]=dep[fa]+1;
	v[dep[x]&1].pb(x);
	for(int i=head[x];i;i=Next[i])
	{
		int u=to[i];
		if(fa==u) continue;
		dfs(u,x);
	}
}
bool vis[M];
int main()
{
/*	freopen("","r",stdin);
	freopen("","w",stdout);*/
	int i;
	cin>>n;
	for(i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y);
		add(y,x);
	}
	dfs(1,1);
	int num=n/3;
	if(v[1].size()<v[0].size())
		swap(v[0],v[1]);
	
	if(v[0].size()<=num)
	{
		for(i=0;i<v[0].size();i++)
		{
			ans[v[0][i]]=(i+1)*3;
			vis[(i+1)*3]=1;
		}
		int tot=1;
		for(i=0;i<v[1].size();i++)
		{
			while(vis[tot]) tot++;
			ans[v[1][i]]=tot;
			vis[tot]=1;
		}
	}
	else 
	{
		int num1=num+(n%3>=1);
		for(i=0;i<num1;i++)
		{
			ans[v[0][i]]=i*3+1;
			vis[i*3+1]=1;
		}
		int j=0;
		for(i;i<v[0].size();i++)
		{
			j++;
			ans[v[0][i]]=j*3;
			vis[j*3]=1;
		}
		int tot=1;
		for(i=0;i<v[1].size();i++)
		{
			while(vis[tot]) tot++;
			ans[v[1][i]]=tot;
			vis[tot]=1;
		}
	}
	for(i=1;i<=n;i++)
		cout<<ans[i]<<" ";
}
/*
7
1 2
2 3
3 4
4 5
5 6
6 7
*/