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;
#define endl '\n'
const int INF=1e9+5;
const int N=2e5+5;
const int mod=998244353;

int id[N];
int n;
vector<int> v[N];
vector<int> have[3];

void dfs(int u,int par){
	for(int& x:v[u]){
		if(x==par) continue;
		id[x]=3-id[u];
		dfs(x,u);
	}
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;
	for(int i=0;i<n-1;++i){
		int x,y;
		cin>>x>>y;
		--x;
		--y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	id[0]=1;
	dfs(0,-1);
	for(int i=0;i<n;++i){
		have[id[i]].push_back(i);
	}
	vector<int> cnt(3);
	for(int i=1;i<=n;++i) ++cnt[i%3];
	vector<int> group(n);
	vector<bool> used(n+1);
	if((int)have[1].size()<=cnt[0]){
		int beg=3;
		for(int x:have[1]){
			group[x]=beg;
			used[beg]=1;
			beg+=3;
		}
		int beg1=1;
		for(int i=0;i<n;++i){
			if(group[i]) continue;
			while(used[beg1]) ++beg1;
			group[i]=beg1;
			++beg1;
		}
	}
	else if((int)have[2].size()<=cnt[0]){
		int beg=3;
		for(int x:have[2]){
			group[x]=beg;
			used[beg]=1;
			beg+=3;
		}
		int beg1=1;
		for(int i=0;i<n;++i){
			if(group[i]) continue;
			while(used[beg1]) ++beg1;
			group[i]=beg1;
			++beg1;
		}
	}
	else{
		int beg=3;
		for(int i=0;i<cnt[1];++i){
			group[have[1][i]]=3*i+1;
		}
		for(int i=cnt[1];i<(int)have[1].size();++i){
			group[have[1][i]]=beg;
			beg+=3;
		}
		for(int i=0;i<cnt[2];++i){
			group[have[2][i]]=3*i+2;
		}
		for(int i=cnt[2];i<(int)have[2].size();++i){
			group[have[2][i]]=beg;
			beg+=3;
		}
	}
	for(int i=0;i<n;++i) cout<<group[i]<<" ";	
	return 0;
}