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>
#define LL long long
#define LD long double
#define pb push_back
#define mp make_pair
#define ss(x) (int) x.size()
#define fi first
#define se second
#define cat(x) cerr << #x << " = " << x << endl
#define rep2(i, j, n) for (LL i = j; i <= n; ++i)
#define rep(i, j, n) for (int i = j; i <= n; ++i)
#define per(i, j, n) for (int i = n; j <= i; --i)
#define boost cin.tie(0);ios_base::sync_with_stdio(0);
#define vi vector <int>

using namespace std;			

const int N = 2e5 + 111;

int n, a, b;
vector <int> v[N];
vector <int> p[2];
vector <int> x[3];

void dfs(int u, int par, int o) {
	o %= 2;
	p[o].pb(u);
	for (auto it : v[u])
		if (it != par)
			dfs(it, u, o + 1);
}

int ans[N];

int main() {	
	scanf ("%d", &n);
	rep(i, 1, n - 1) {
		scanf ("%d%d", &a, &b);
		v[a].pb(b);
		v[b].pb(a);
	}
	dfs(1, 0, 0);
	rep(i, 1, n) 
		x[i % 3].pb(i);
	if (ss(p[1]) > ss(p[0])) swap(p[0], p[1]);

	while (ss(x[1])) {
		ans[p[0].back()] = x[1].back();
		p[0].pop_back();
		x[1].pop_back();
	}
	
	if (ss(p[1]) > ss(p[0])) swap(p[0], p[1]);
	
	while (ss(x[2])) {
		ans[p[0].back()] = x[2].back();
		p[0].pop_back();
		x[2].pop_back();
	}
	
	while (ss(x[0])) {
		if (ss(p[0])) {
			ans[p[0].back()] = x[0].back();
			p[0].pop_back();
		}
		else {
			ans[p[1].back()] = x[0].back();
			p[1].pop_back();
		}
		x[0].pop_back();
	}
	rep(i, 1, n) printf ("%d ", ans[i]);
		
	
	
	
	
	return 0;
}