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 <vector>
using namespace std;
const int maxNode = 2e5 + 5;
vector<int> tree[maxNode];
int nbNode, ans[maxNode];
vector<int> col[2], kind[3];
void Dfs(int now, int from, int faCol) {
	col[faCol ^ 1].push_back(now);
	for (int to : tree[now]) {
		if (to == from) continue ;
		Dfs(to, now, faCol ^ 1);
	}
}
int main() {
	scanf("%d", &nbNode);
	for (int edge = 1; edge < nbNode; ++edge) {
		int u, v;
		scanf("%d%d", &u, &v);
		tree[u].push_back(v);
		tree[v].push_back(u);
	}
	for (int num = 1; num <= nbNode; ++num)
		kind[num % 3].push_back(num);
	Dfs(1, -1, 1);
	if (col[0].size() > col[1].size())
		swap(col[0], col[1]);
	if (col[0].size() <= kind[0].size()) {
		int now = 0, pt = 0;
		for (int node : col[0])
			ans[node] = kind[0][pt++];
		if (pt >= kind[0].size()) {
			++now;
			pt = 0;
		}
		for (int node = 1; node <= nbNode; ++node)
			if (!ans[node]) {
				ans[node] = kind[now][pt++];
				if (pt >= kind[now].size()) {
					pt = 0;
					++now;
				}
			}
	} else {
		int pt = 0;
		for (int num : kind[1])
			ans[col[0][pt]] = num, ++pt;
		pt = 0;
		for (int num : kind[2])
			ans[col[1][pt]] = num, ++pt;
		pt = 0;
		for (int node = 1; node <= nbNode; ++node)
			if (!ans[node])
				ans[node] = kind[0][pt++];
	}
	for (int node = 1; node <= nbNode; ++node)
		printf("%d ", ans[node]);
	printf("\n");
	return 0;
}