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 maxn=200100;
int n;
vector<int> ver[maxn],A,B,T;
inline void dfs(int x,int fat,int v)
{
	if(v) A.push_back(x);else B.push_back(x);
	for(int i=0;i<(int)ver[x].size();i++)
	{
		int y=ver[x][i];if(y==fat) continue;
		dfs(y,x,v^1);
	}
}
int vis[maxn],ans[maxn];
int main()
{
	cin>>n;
	for(int i=1,a,b;i<n;i++)
	{
		scanf("%d%d",&a,&b);
		ver[a].push_back(b);
		ver[b].push_back(a);
	}
	dfs(1,0,0);
	if(A.size()>B.size()) swap(A,B);
	if(A.size()<=n/3)
	{
		for(int i=0;i<(int)A.size();i++) vis[i*3+3]=1,ans[A[i]]=i*3+3;
		for(int i=1;i<=n;i++) if(!vis[i]) T.push_back(i);
		for(int i=0;i<(int)B.size();i++) ans[B[i]]=T[i];
	}
	else
	{
		int nowa=1,nowb=2,nowc=3;
		for(int i=0;i<(int)A.size();i++) if(nowa<=n) ans[A[i]]=nowa,nowa+=3;else ans[A[i]]=nowc,nowc+=3;
		for(int i=0;i<(int)B.size();i++) if(nowb<=n) ans[B[i]]=nowb,nowb+=3;else ans[B[i]]=nowc,nowc+=3;
	}
	for(int i=1;i<=n;i++) printf("%d ",ans[i]);
	return 0;
}