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:
//%std
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
	int out = 0, fh = 1;
	char jp = getchar();
	while ((jp > '9' || jp < '0') && jp != '-')
		jp = getchar();
	if (jp == '-')
		fh = -1, jp = getchar();
	while (jp >= '0' && jp <= '9')
		out = out * 10 + jp - '0', jp = getchar();
	return out * fh;
}
const int N = 2e5 + 10;
vector<int> G[N];
int n, t[2], k[3], dep[N], ans[N];
void dfs(int u, int F, int d)
{
	t[d]++;
	dep[u] = d;
	for (int v : G[u])
		if (v != F)
			dfs(v, u, d ^ 1);
}
int main()
{
	n = read();
	for (int i = 1; i < n; ++i)
	{
		k[i % 3]++;
		int u = read(), v = read();
		G[u].push_back(v);
		G[v].push_back(u);
	}
	k[n % 3]++;
	dfs(1, 0, 0);
	int p = 0;
	if (t[0] > t[1])
		swap(t[0], t[1]), p = 1;
	if (k[0] >= t[0])
	{
		for (int i = 1; i <= n; ++i)
			if (dep[i] == p)
				ans[i] = k[0]-- * 3;
		for (int i = 1; i <= n; ++i)
			if (dep[i] != p)
			{
				if (k[0])
					ans[i] = k[0]-- * 3;
				else if (k[1])
					ans[i] = (k[1]-- - 1) * 3 + 1;
				else
					ans[i] = (k[2]-- - 1) * 3 + 2;
			}
	}
	else if (k[1] <= t[0])
	{
		for (int i = 1; i <= n; ++i)
			if (dep[i] == p)
			{
				if (k[1])
					ans[i] = (k[1]-- - 1) * 3 + 1;
				else
					ans[i] = k[0]-- * 3;
			}
		for (int i = 1; i <= n; ++i)
			if (dep[i] != p)
			{
				if (k[0])
					ans[i] = k[0]-- * 3;
				else if (k[1])
					ans[i] = (k[1]-- - 1) * 3 + 1;
				else
					ans[i] = (k[2]-- - 1) * 3 + 2;
			}
	}
	else
		assert(false);
	for (int i = 1; i <= n; ++i)
		printf("%d ", ans[i]);
	puts("");
	return 0;
}