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 = 200010;

int n;
vector<int> G[N], v[2];
int p[N];

void dfs(int u, int fa, bool f) {
	v[f].push_back(u);
	for(auto &i : G[u]) if(i != fa) dfs(i, u, f ^ 1);
} 

int main() {
	scanf("%d", &n);
	for(int u, v, i = 1; i < n; ++i) {
		scanf("%d%d", &u, &v);
		G[v].push_back(u);
		G[u].push_back(v);
	}
	dfs(1, 1, 1);
	if(v[0].size() > v[1].size()) swap(v[0], v[1]);
	int x = n / 3;
	if((int)v[0].size() <= x) {
		int now = 3, T = 0;
		for(auto &i : v[0]) p[i] = now, now += 3;
		for(auto &i : v[1]) {
			if(now > n) now = ++T;
			p[i] = now;
			now += 3;
		}
	} else {
		int now = 1;
		for(auto &i : v[0]) {
			p[i] = now;
			now += 3;
			if(now > n) now = 3;
		}
		int T = now;
		now = 2;
		for(auto &i : v[1]) {
			p[i] = now;
			now += 3;
			if(now > n) now = T;
		}
	}
	for(int i = 1; i <= n; ++i) printf("%d ", p[i]);
	return 0;
}