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<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
const int MAXN = 2e5 + 5;

struct Edge
{
	int next,to;
}e[MAXN<<1];
int head[MAXN],etot=0;
inline void add(int u,int v)
{
	e[++etot]=(Edge){ head[u],v};
	head[u]=etot;
}

int clr[MAXN],ccnt[3];
void dfs(int u,int fa)
{
	++ccnt[clr[u]];
	for(int i=head[u]; i; i=e[i].next)
	{
		int v=e[i].to;
		if(v==fa) continue;
		clr[v]=!clr[u];
		dfs(v,u);
	}
}

queue<int> q[4];

int main(void)
{
	int n;
	scanf("%d",&n);
	for(int i=1; i<n; ++i)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v); add(v,u);
	}
	
	dfs(1,0);
	
	for(int i=1; i<=n; ++i)
		q[i%3].push(i);
	int c1 = q[1].size();
	int c2 = q[2].size();
	
	if(c1 <= ccnt[0] && c2 <= ccnt[1])
		for(int i=1; i<=n; ++i)
		{
			int x = clr[i]+1;
			if(q[x].empty()) x=0;
			printf("%d ",q[x].front());
			q[x].pop();
		}
	else if(c1 <= ccnt[1] && c2 <= ccnt[0])
		for(int i=1; i<=n; ++i)
		{
			int x = (!clr[i])+1;
			if(q[x].empty()) x=0;
			printf("%d ",q[x].front());
			q[x].pop();
		}
	else if(c1+c2 <= ccnt[0])
		for(int i=1; i<=n; ++i)
		{
			int x;
			if(clr[i]) x=0;
			else if(!q[1].empty()) x=1;
			else if(!q[2].empty()) x=2;
			else x=0;
			printf("%d ",q[x].front());
			q[x].pop();
		}
	else if(c1+c2 <= ccnt[1])
		for(int i=1; i<=n; ++i)
		{
			int x;
			if(!clr[i]) x=0;
			else if(!q[1].empty()) x=1;
			else if(!q[2].empty()) x=2;
			else x=0;
			printf("%d ",q[x].front());
			q[x].pop();
		}
	else printf("-1");
	return 0;
}