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;

#define ll long long
#define mp make_pair
#define pb push_back

const int N = 2e5;
int n;
vector <int> g[N];
int col[N], cnt[2];
int ans[N];
int used[N];

void dfs(int u, int p, int c) {
	col[u] = c; cnt[c] ++;
	for (int i = 0; i < g[u].size(); ++ i) {
		int v = g[u][i];
		if (v != p) {
			dfs(v, u, c ^ 1);
		}
	}
}

int main() {
	ios :: sync_with_stdio(0);
	cin >> n;
	for (int i = 1; i < n; ++ i) {
		int u, v; cin >> u >> v;
		-- u; -- v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(0, -1, 0);
	if (cnt[0] > cnt[1]) {
		swap(cnt[0], cnt[1]);
		for (int i = 0; i < n; ++ i) col[i] ^= 1;
	}
	int x = n / 3;
	if (cnt[0] <= x) {
		int nw = 0;
		for (int i = 0; i < n; ++ i) if (!col[i]) used[ans[i] = (nw += 3)] = 1;
		vector <int> v;
		for (int i = 0; i < n; ++ i) if (col[i]) v.push_back(i);
		for (int i = 1, j = 0; i <= n; ++ i) if (!used[i]) ans[v[j ++]] = i;
	} else {
		int nw0 = 1, nw1 = 2, nw2 = 0;
		for (int i = 0; i < n; ++ i) {
			if (!col[i]) {
				if (nw0 <= n) ans[i] = nw0, nw0 += 3;
				else ans[i] = (nw2 += 3);
			} else {
				if (nw1 <= n) ans[i] = nw1, nw1 += 3;
				else ans[i] = (nw2 += 3);
			}
		}
	}
	for (int i = 0; i < n; ++ i) cout << ans[i] << " \n"[i + 1 == n];
	return 0;
}