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 <iostream>
#include <cstdio>
#include <cstring>
#include <cctype>
#define MAXN 200005
#define MAXM 400005
using namespace std;
struct edge{int u,v;}e[MAXM];
int head[MAXN],nxt[MAXM],cnt;
void addnode(int u,int v)
{
	e[++cnt]=(edge){u,v};
	nxt[cnt]=head[u];
	head[u]=cnt;
}
int col[MAXN],tot[2],p[MAXN],n; 
void dfs(int u)
{
	++tot[col[u]];
	for (int i=head[u];i;i=nxt[i])
		if (col[e[i].v]==-1)
			col[e[i].v]=col[u]^1,dfs(e[i].v);
}
inline void putp(int r,int c)//freedomly put r+3k into color c
{
	int cur=r;
	for (int i=1;i<=n;i++)
		if (col[i]==c&&!p[i])
		{
			p[i]=cur,cur+=3;
			if (cur>n) return;
		}
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		addnode(u,v),addnode(v,u);
	}
	memset(col,-1,sizeof(col));
	for (int i=1;i<=n;i++) if (col[i]==-1) col[i]=0,dfs(i);
	int k=(n+1)/3;
	if (k<=min(tot[0],tot[1])) putp(2,tot[1]<=tot[0]);
	else putp(2,tot[1]>tot[0]);
	putp(1,tot[1]>tot[0]);
	int cur=3;
	for (int i=1;i<=n;i++)
		if (!p[i])
			p[i]=cur,cur+=3;
	for (int i=1;i<=n;i++) printf("%d%c",p[i]," \n"[i==n]);
	return 0;
}