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 MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
vector<int> G[200005];
int col[200005];
int cnt[2];
int ans[200005];

void dfs(int v,int c){
	col[v]=c;
	for(int i=0;i<G[v].size();i++){
		if(col[G[v][i]]==-1){
			dfs(G[v][i],1-c);
		}
	}
}

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		a--;
		b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	memset(col,-1,sizeof(col));
	dfs(0,0);
	for(int i=0;i<n;i++){
		cnt[col[i]]++;
	}
	if(cnt[0]<cnt[1]){
		swap(cnt[0],cnt[1]);
		for(int i=0;i<n;i++){
			col[i]=1-col[i];
		}
	}
	int th=n/3;
	int fi=(n+2)/3;
	int sc=(n+1)/3;
	int ma=cnt[0];
	int mi=cnt[1];
	if(ma>th+fi && mi>th){
		assert(false);
		printf("-1\n");
		return 0;
	}
	int cur=1;
	int cur2=2;
	int cur3=3;
	if(mi<=th){
		for(int i=0;i<n;i++){
			if(col[i]==0){
				if(fi>0){
					fi--;
					ans[i]=cur;
					cur+=3;
				}else if(sc>0){
					sc--;
					ans[i]=cur2;
					cur2+=3;
				}else if(fi==0 && sc==0){
					ans[i]=cur3;
					cur3+=3;
				}
			}else{
				ans[i]=cur3;
				cur3+=3;
			}
		}
	}else{
		for(int i=0;i<n;i++){
			if(col[i]==0){
				if(fi>0){
					fi--;
					ans[i]=cur;
					cur+=3;
				}else{
					ans[i]=cur3;
					cur3+=3;
				}
			}else{
				if(sc>0){
					sc--;
					ans[i]=cur2;
					cur2+=3;
				}else{
					ans[i]=cur3;
					cur3+=3;
				}
			}
		}
	}
	for(int i=0;i<n;i++){
		printf("%d%c",ans[i],i+1==n?'\n':' ');
	}
	return 0;
}