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 eb emplace_back
#define all(v) v.begin(),v.end()
using namespace std;
using ll=long long;
using P=pair<ll,ll>;
const ll INF=1LL<<60;
const ll MOD=998244353LL;
template<class T> void chmin(T &a,T b){a=min(a,b);}
template<class T> void chmax(T &a,T b){a=max(a,b);}
vector<int> dep,c;
vector<vector<int>> G;
void dfs(int i,int p){
	if(dep[i]%2==0){
		c[i]=1;
	}else{
		c[i]=2;
	}
	for(auto &e:G[i]){
		if(e==p)continue;
		dep[e]=dep[i]+1;
		dfs(e,i);
	}
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;cin>>n;
	G.resize(n);
	for(int i=0;i<n-1;i++){
		int a,b;cin>>a>>b;--a;--b;
		G[a].eb(b);
		G[b].eb(a);
	}
	dep.resize(n);
	c.resize(n);
	dfs(0,-1);
	int c0=0,c1=0,c2=0;
	int cc0=n/3,cc1=(n+2)/3,cc2=(n+1)/3;
	for(int i=0;i<n;i++){
		if(c[i]==1){
			c1++;
		}else{
			c2++;
		}
	}
	if(c1>c2){
		for(int i=0;i<n;i++){
			if(c[i]==1){
				c[i]=2;
			}else{
				c[i]=1;
			}
		}
		swap(c1,c2);
	}
	if(c1<cc1){
		for(int i=0;i<n;i++){
			if(c[i]==1){
				c[i]=0;
			}
		}
		c0=c1;
		c1=0;
		for(int i=0;i<n;i++){
			if(c1<cc1&&c[i]==2){
				c1++;
				c[i]=1;
				c2--;
			}
		}
	}
	for(int i=0;i<n;i++){
		if(c0<cc0){
			if(c2>cc2&&c[i]==2){
				c2--;
				c0++;
				c[i]=0;
			}
			if(c1>cc1&&c[i]==1){
				c1--;
				c0++;
				c[i]=0;
			}
		}
	}
	int t0=3,t1=1,t2=2;
	for(int i=0;i<n;i++){
		if(c[i]==0){
			cout<<t0<<" ";
			t0+=3;
		}else if(c[i]==1){
			cout<<t1<<" ";
			t1+=3;
		}else{
			cout<<t2<<" ";
			t2+=3;
		}
	}
	cout<<endl;
}