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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
#define N 200010
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,a[N],dep[N],cnt0,cnt1,res1,res2;
vector<int> vec[N],G[N];
void dfs(int u,int fa){
	dep[u]=dep[fa]+1;
	vec[dep[u]].push_back(u);
	for(auto v:G[u]){
		if(v==fa)continue;
		dfs(v,u);
	}
}
int main(){
	n=read();
	for(int i=1;i<n;++i){
		int u=read(),v=read();
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(1,0);
	for(int i=1;i<=n;++i){
		if(i&1)cnt1+=vec[i].size();
		else cnt0+=vec[i].size();
	}
	res1=n/3+(n%3>=1),res2=n/3+(n%3>=2);
	if(max(cnt1,cnt0)>=res1&&min(cnt0,cnt1)>=res2){
		for(int i=1;i<=n;++i){
			if((i&1)^(cnt1<=cnt0)){
				for(auto u:vec[i]){
					if(!res1)break;
					--res1;
					a[u]=res1*3+1;
				}
			}
			else{
				for(auto u:vec[i]){
					if(!res2)break;
					--res2;
					a[u]=res2*3+2;
				}
			}
		}
	}
	else{
		for(int i=1+(cnt1<cnt0);i<=n;i+=2){
			for(auto u:vec[i]){
				if(res1){
					--res1;
					a[u]=res1*3+1;
				}
				else if(res2){
					--res2;
					a[u]=res2*3+2;
				}
			}
		}
	}
	int now=1;
	for(int i=1;i<=n;++i){
		if(!a[i]){
			a[i]=now*3;
			++now;
		}
	}
	for(int i=1;i<=n;++i){
		printf("%d ",a[i]);
	}
	printf("\n");
	return 0;
}