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;

typedef long long ll;

int n;
vector<int> G[200010];

int col[200010];
void dfs(int v,int p,int c){
	col[v]=c;
	for(int u: G[v]){
		if(u==p)continue;
		dfs(u,v,c^1);
	}
}

int main(){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		G[a].push_back(b);
		G[b].push_back(a);
	}
	
	dfs(1,0,0);
	int cnt[2]={};
	for(int i=1;i<=n;i++){
		cnt[col[i]]++;
	}
	int a,b,c;
	a=b=c=n/3;
	if(n%3>=1)a++;
	if(n%3>=2)b++;
	
	static int ret[200010];
	static bool used[200010];
	for(int i=0;i<200010;i++){
		used[i]=false;
	}
	if(cnt[0]<=c||cnt[1]<=c){
		int t=0; if(cnt[0]>c)t=1;
		int CNT=0;
		for(int i=1;i<=n;i++){
			if(col[i]==t){
				CNT++;
				ret[i]=CNT*3;
				used[ret[i]]=true;
			}
		}
		CNT=1;
		for(int i=1;i<=n;i++){
			if(col[i]!=t){
				while(used[CNT])CNT++;
				ret[i]=CNT;
				used[ret[i]]=true;
			}
		}
	}
	else {
		cnt[0]=cnt[1]=0;
		int k=3;
		for(int i=1;i<=n;i++){
			if(3*cnt[col[i]]+1+col[i]<=n){
				ret[i]=3*cnt[col[i]]+1+col[i];
			}
			else {
				while(used[k])k+=3;
				ret[i]=k;
			}
			cnt[col[i]]++;
			used[ret[i]]=true;
		}
	}
	
	for(int i=1;i<=n;i++){
		printf("%d%c",ret[i],(i==n)?'\n':' ');
	}
}