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 pb push_back
#define mp make_pair
#define fst first
#define snd second
using namespace std;

typedef long long ll;
typedef double db;
typedef long double ldb;

template<class T> inline void checkmin(T &a,const T &b){if(b<a) a=b;}
template<class T> inline void checkmax(T &a,const T &b){if(b>a) a=b;}

const int inf=1e9+10;
const int maxn=2e5+10;

vector <int> g[maxn],v[3],c[2];
int n,ans[maxn];

void dfs(int x,int fa,int col){
	c[col].pb(x);
	for(int i=0;i<g[x].size();i++){
		int to=g[x][i];
		if(to!=fa) dfs(to,x,col^1);
	}
}

int main(){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int u,vv;
		scanf("%d%d",&u,&vv);
		u--; vv--;
		g[u].pb(vv);
		g[vv].pb(u);
	}
	dfs(0,-1,0);
	for(int i=1;i<=n;i++) v[i%3].pb(i);
	if(c[0].size()>c[1].size()) swap(c[0],c[1]);
	vector <int> v2,v3;
	if(c[0].size()>=v[2].size()){
		for(int i=0;i<v[2].size();i++)
			v2.pb(v[2][i]);
		for(int i=0;i<v[0].size();i++)
			v2.pb(v[0][i]);
		for(int i=0;i<v[1].size();i++)
			v2.pb(v[1][i]);
	}
	else{
		for(int i=0;i<v[0].size();i++)
			v2.pb(v[0][i]);
		for(int i=0;i<v[1].size();i++)
			v2.pb(v[1][i]);
		for(int i=0;i<v[2].size();i++)
			v2.pb(v[2][i]);
	}
	for(int i=0;i<c[0].size();i++) v3.pb(c[0][i]);
	for(int i=0;i<c[1].size();i++) v3.pb(c[1][i]);
	for(int i=0;i<n;i++)
		ans[v3[i]]=v2[i];
	for(int i=0;i<n;i++) printf("%d ",ans[i]);
	printf("\n");
	return 0;
}