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;
const int maxn=2e5+7;
vector<int> g[maxn];
int co[maxn];
int ans[maxn];
int use[maxn]; 
vector<int> r;
vector<int> b;
void dfs(int now,int x){
	co[now]=x;
	for(int i=0;i<g[now].size();i++){
		int u=g[now][i];
		if(co[u]==0)  dfs(u,x*-1);
	}
}

int main(){
	int n;
	cin>>n;
	int t=n-1;
	while(t--){
		int a,b;
		cin>>a>>b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	dfs(1,1);
	for(int i=1;i<=n;i++){
		if(co[i]==1) r.push_back(i);
		if(co[i]==-1) b.push_back(i);
	}
	if(r.size()>b.size())	swap(r,b);
	if(r.size()>n/3){
		int i=0;
		for(int k=0;3*k+1<=n;k++){
			ans[r[i++]]=3*k+1;
			use[3*k+1]=1;
		}
		i=0;
		for(int k=0;3*k+2<=n;k++){
			ans[b[i++]]=3*k+2;
			use[3*k+2]=1;
		}
		int k=1;
		for(int i=1;i<=n;i++){
			if(ans[i]==0){
				ans[i]=3*k;k++;
			}
		}
	}
	else{
		int k=1;
		for(int i=0;i<r.size();i++){
			ans[r[i]]=3*k;
			use[3*k]=1;
			k++;
		}
		k=1;
		for(int i=1;i<=n;i++){
			if(ans[i]==0){
				while(use[k]==1){
					k++;
				}
				ans[i]=k;
				use[k]=1;
			}
		}
	}
	for(int i=1;i<=n;i++) cout<<ans[i]<<" ";
	return 0;
}