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 int lld;

vector<int> v[222222],dp1,dp2;
int f,s,ans[222222];
void recur(int now,int left,int dep){
	if(dep%2==0) dp1.push_back(now);
	else dp2.push_back(now);
	for(int e=0;e<v[now].size();e++){
		int next = v[now][e];
		if(next!=left) recur(next,now,dep+1);
	}
}
int main(void){
	int n;
	scanf("%d",&n);
	for(int e=0;e<n-1;e++){
		int a,b;
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	recur(1,-1,0);
	if((int)dp1.size()>(int)dp2.size()) swap(dp1,dp2);
	if((int)dp1.size()<=(n/3)){
		int q = 3;
		for(int p=0;p<dp1.size();p++){
			 ans[dp1[p]] = q;
			 q += 3;
		}
		q -= 3;
		int tmp = 1;
		for(int e=1;e<=n;e++){
			if(ans[e]==0){
				ans[e] = tmp;
				tmp++;
				if(tmp%3==0&&tmp<=q) tmp++;
			}
		}
		for(int e=1;e<=n;e++) printf("%d ",ans[e]);
	}else{
		for(int q=0,tmp=1;q<dp1.size()&&tmp<=n;q++,tmp+=3){
			ans[dp1[q]] = tmp;
		}
		for(int q=0,tmp=2;q<dp2.size()&&tmp<=n;q++,tmp+=3){
			ans[dp2[q]] = tmp;
		}
		int tmp = 3;
		for(int e=1;e<=n;e++){
			if(ans[e]==0){
				ans[e] = tmp;
				tmp += 3;
			}
		}
		for(int e=1;e<=n;e++) printf("%d ",ans[e]);
	}
	return 0;
}