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 N 200100
using namespace std;
vector<int> g[N];
vector<int> v[3];
int c[N],ans[N];
bool used[N];
void DFS(int u,int cc){
	v[cc].push_back(u);
	c[u]=cc;
	for(int i=0;i<g[u].size();i++){
		if(!c[g[u][i]]) DFS(g[u][i],3-cc);
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		g[x].push_back(y);
		g[y].push_back(x);
	}
	DFS(1,1);
	int mc=1,cur=1;
	if(v[2].size()<v[1].size()) mc=2;
	if(n/3>=v[mc].size()){
		for(int i=0;i<v[mc].size();i++){
			ans[v[mc][i]]=(i+1)*3;
			used[(i+1)*3]=true;
		}
		for(int i=0;i<v[3-mc].size();i++){
			while(used[cur]) cur++;
			used[cur]=true;
			ans[v[3-mc][i]]=cur;
		}
	}
	else{
		for(int j=1;j<=2;j++){
			for(int k=j;k<=n;k+=3){
				ans[v[j][k/3]]=k;
				used[v[j][k/3]]=true;
			}
		}
		for(int j=3;j<=n;j+=3){
			while(used[cur]) cur++;
			used[cur]=true;
			ans[cur]=j;
		}
	}
	for(int i=1;i<=n;i++) printf("%d ",ans[i]);
	return 0;
}