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 = 5e5 + 10;

struct Node
{
	int to, next;
}edge[maxN * 2 + 1];

int head[maxN + 1], n, tot;
int c0, c1, c2, ans[maxN + 1];

vector<int> num[2];

inline void add(int x, int y)
{
	tot ++;
	edge[tot].to = y;
	edge[tot].next = head[x];
	head[x] = tot;
}

inline void dfs(int u, int pa, int dep)
{
	num[dep].push_back(u);
	for(int i = head[u]; i; i = edge[i].next)
	{
		int v = edge[i].to;
		if(v == pa) continue;
		dfs(v, u, dep ^ 1);
	}
}

int main()
{
	scanf("%d", &n);
	c0 = n / 3, c1 = (n + 2) / 3, c2 = (n + 1) / 3;

	for(int i = 1; i < n; i++)
	{
		int x, y;
		scanf("%d %d", &x, &y);
		add(x, y); add(y, x);
	}

	dfs(1, 0, 1);

	if(num[0].size() > num[1].size()) swap(num[0], num[1]);

	if(num[0].size() <= c0)
	{
		for(int i = 0; i < num[0].size(); i++)
			ans[ num[0][i] ] = 0, c0 --;
	}
	else
	{
		for(int i = 0; i < num[0].size(); i++)
			if(c1) ans[ num[0][i] ] = 1, c1 --;
			else if(c0) ans[ num[0][i] ] = 0, c0 --;
	}

	for(int i = 0; i < num[1].size(); i++)
		if(c1) ans[ num[1][i] ] = 1, c1 --;
		else if(c2) ans[ num[1][i] ] = 2, c2 --;
			else if(c0) ans[ num[1][i] ] = 0, c0 --;

	int t[3]; t[1] = t[2] = 0; t[0] = 1;
	for(int i = 1; i <= n; i++)
		printf("%d ", t[ ans[i] ] * 3 + ans[i]), t[ ans[i] ] ++;
	return 0;
}