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:
/*
Date&Time: 12:19pm 3/10/2020
*/

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 2e5 + 10;

int ver[N * 2], nex[N * 2], head[N], cnt[2], ans[N], color[N];
int n, tot;
vector<int> a[3];

inline void add(int x, int y) {
	ver[++tot] = y;
	nex[tot] = head[x];
	head[x] = tot;
}

void dfs(bool c, int x, int fa) {
	color[x] = c;
	cnt[c]++;
	for (int i = head[x]; i; i = nex[i]) {
		int y = ver[i];
		if (y == fa) continue;
		dfs(!c, y, x);
	}
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) {
		a[i % 3].push_back(i);
	}
	for (int i = 1, u, v; i < n; i++) {
		cin >> u >> v;
		add(u, v), add(v, u);
	}
	dfs(1, 1, 0);
	int fl = 0;
	int r = cnt[0], b = cnt[1];
	if (r > b) swap(r, b), fl = !fl;
	int x = n / 3;
	int zero = 0, one = 0, two = 0;
	if (r <= x) {
		for (int i = 1; i <= n; i++) {
			if (color[i] == fl) {
				ans[i] = a[0][zero++];
			}
			else {
				ans[i] = one < a[1].size() ? a[1][one++] : two < a[2].size() ? a[2][two++] : a[0][zero++];
			}
		}
	}
	else{
		for (int i = 1; i <= n; i++) {
			if (color[i] == fl) {
				ans[i] = one < a[1].size() ? a[1][one++] : zero < a[0].size() ? a[0][zero++] : a[1][one++];
			}
			else {
				ans[i] = two < a[2].size() ? a[2][two++] : zero < a[0].size() ? a[0][zero++] : a[1][one++];
			}
		}
	}
	for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
	cout << '\n';
}