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;
const int N=2e5+5;
vector<int>v[N];
int d[N],num[2];
void dfs(int x,int fa){
	for(auto u:v[x]){
		if(u==fa)continue;
		d[u]=d[x]^1,dfs(u,x);
	}
	num[d[x]]++;
}
int main(){
	int n,x,y;scanf("%d",&n);
	for(int i=1;i<n;i++)scanf("%d%d",&x,&y),v[x].push_back(y),v[y].push_back(x);
	dfs(1,0);
	int a=1,b=2,c=3;
	if(num[0]<=n/3){
		for(int i=1;i<=n;i++){
			if(d[i]%2==0)printf("%d ",c),c+=3;
			else printf("%d ",a<=n?(a+=3)-3:(b<=n?(b+=3)-3:(c+=3)-3));
		}
		return 0;
	}
	if(num[1]<=n/3){
		for(int i=1;i<=n;i++){
			if(d[i]&1)printf("%d ",c),c+=3;
			else printf("%d ",a<=n?(a+=3)-3:(b<=n?(b+=3)-3:(c+=3)-3));
		}
		return 0;
	}
	for(int i=1;i<=n;i++){
		if(d[i]&1)printf("%d ",a<=n?(a+=3)-3:(c+=3)-3);
		else printf("%d ",b<=n?(b+=3)-3:(c+=3)-3);
	}
	return 0;
}