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 <vector>
using namespace std;
#pragma warning (disable: 4996)

int N, A[1 << 18], B[1 << 18];
int col[1 << 18];
vector<int> X[1 << 18];
vector<int> L0, L1;

int P[1 << 18];
bool used[1 << 18];

void dfs(int pos, int dep) {
	col[pos] = dep;
	if (dep == 0) L0.push_back(pos);
	else L1.push_back(pos);
	for (int i : X[pos]) {
		if (col[i] >= 0) continue;
		dfs(i, 1 ^ dep);
	}
}

int main() {
	cin >> N;
	for (int i = 1; i <= N - 1; i++) {
		cin >> A[i] >> B[i];
		X[A[i]].push_back(B[i]);
		X[B[i]].push_back(A[i]);
	}
	for (int i = 1; i <= N; i++) col[i] = -1;
	dfs(1, 0);
	if (L0.size() > L1.size()) swap(L0, L1);

	int num = N / 3;
	if (L0.size() <= num) {
		for (int i = 0; i < L0.size(); i++) {
			P[L0[i]] = 3 * (i + 1);
			used[3 * (i + 1)] = true;
		}
		int cnts = 1;
		for (int i = 0; i < L1.size(); i++) {
			while (used[cnts] == true) cnts++;
			used[cnts] = true;
			P[L1[i]] = cnts;
		}
	}
	else {
		for (int i = 0; i < L0.size(); i++) {
			if (3 * i + 1 > N) break;
			P[L0[i]] = 3 * i + 1;
			used[3 * i + 1] = true;
		}
		for (int i = 0; i < L1.size(); i++) {
			if (3 * i + 2 > N) break;
			P[L1[i]] = 3 * i + 2;
			used[3 * i + 2] = true;
		}
		int cnts = 1;
		for (int i = 1; i <= N; i++) {
			if (P[i] >= 1) continue;
			while (used[cnts] == true) cnts++;
			used[cnts] = true;
			P[i] = cnts;
		}
	}

	for (int i = 1; i <= N; i++) {
		if (i >= 2) cout << " ";
		cout << P[i];
	}
	cout << endl;
	return 0;
}