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;
vector<int> E[200010];
int P[200010]; 
int cnt[2]; 
int ans[200010]; 
bool used[200010]; 
void dfs(int i, int p, int n){
	P[i] = n; 
	cnt[n] ++; 
	for(auto v : E[i]){
		if(v == p) continue; 
		dfs(v, i, 1-n); 
	}
}
int main(){
	int N; cin >> N; 
	for(int i = 0; i < N-1; i++){
		int a, b; cin >> a >> b; 
		a--; b--; 
		E[a].push_back(b); E[b].push_back(a); 
	}
	dfs(0, -1, 0); 
	int mn = min(cnt[0], cnt[1]); 
	if(cnt[0] > cnt[1]){
		for(int i = 0; i < N; i++) P[i] = 1 - P[i]; 
	}
	if(mn <= N / 3){
		int now = 3; 
		for(int i = 0; i < N; i++){
			if(P[i] == 0){
				ans[i] = now; now += 3; 
				used[ans[i]] = true; 
			}
		}
		now = 1; 
		for(int i = 0; i < N; i++){
			if(ans[i]) continue; 
			while(used[now]) now ++; 
			ans[i] = now ++; 
		}
	}else{
		int now1 = 1; int now2 = 2; 
		for(int i = 0; i < N; i++){
			if(P[i] == 0 && now2 <= N){
				ans[i] = now2; now2 += 3; 
				used[ans[i]] = true; 
			}
			if(P[i] == 1 && now1 <= N){
				ans[i] = now1; now1 += 3; 
				used[ans[i]] = true; 
			}
		}
		int now = 1; 
		for(int i = 0; i < N; i++){
			if(ans[i]) continue; 
			while(used[now]) now ++; 
			ans[i] = now ++; 
		}
	}
	for(int i = 0; i < N; i++) cout << ans[i] << " "; 
	cout << "\n"; 
}