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;
typedef long long ll;

const int N = 2e5;

int cnt[3];
int col[N + 5];
int ans[N + 5];
vector < int > G[N + 5];

void dfs(int u, int p, int c) {
	cnt[c]++;
	col[u] = c;
	for(int v : G[u]) {
		if(v == p) continue;
		dfs(v, u, 1^c);
	}
}

int main() {
	ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
	memset(ans,-1,sizeof ans);
	int n; cin >> n;
	for(int i=0; i<n-1; i++) {
		int u,v; cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(1,-1,0);
//	for(int i=1; i<=n; i++) printf("%d ",col[i]);
//	puts("");
	int val[5], rem[5];
	for(int i=1; i<=3; i++) {
		rem[i] = (3-i+n)/3;
		val[i] = i;
	}
	int id[3];
	id[0] = 0; id[1] = 1;
	if(cnt[0] > cnt[1]) {
		swap(id[0], id[1]);
	}
	for(int i=0; i<2; i++) {
		if(cnt[id[i]] <= rem[3]) {
			for(int j=1; j<=n; j++) {
				if(col[j] == id[i]) {
					ans[j] = val[3];
					val[3] += 3;
					rem[3]--;
				}
			}
		}
	}
	
	for(int i=0; i<2; i++) {
		for(int j=1; j<=n; j++) {
			if(ans[j] != -1) continue;
			if(rem[id[i]+1] == 0) break;
			if(col[j] == id[i]) {
				ans[j] = val[id[i]+1];
				val[id[i]+1] += 3;
				rem[id[i]+1]--;
			}
		}
	}
	for(int i=1; i<=n; i++) {
		if(ans[i] != -1) continue;
		for(int j=3; j>=1; j--) {
			if(rem[j] == 0) continue;
			ans[i] = val[j];
			val[j] += 3;
			rem[j]--;
			break;
		}
	}
	for(int i=1; i<=n; i++) {
		if(i > 1) cout << " ";
		cout << ans[i];
	}
	cout << endl;
	return 0;
}