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;

const int maxn = 2e5 + 10;

int cnt[2], ans[maxn];
vector<int> g[maxn], p[2];

void dfs(int o,int d,int pa = 0) {
	cnt[d&1] += 1;
	p[d&1].push_back(o);
	for(auto u : g[o]) if(u != pa) dfs(u, d + 1, o);
}
int main() {
	int n;
	scanf("%d", &n);
	for(int i = 1;i < n;++i) {
		int a , b;
		scanf("%d%d", &a, &b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	dfs(1, 0);
	int a = 0,b = 1;
	if(cnt[a] > cnt[b]) swap(a, b);
	int o = 1, t = 2, z = 3;
	if(cnt[a] <= (n / 3)) {
		for(auto u : p[a]) {
			if(z <= n) ans[u] = z, z += 3;
		}
		for(auto u : p[b]) {
			if(z <= n) ans[u] = z, z += 3;
			else if(o <= n) ans[u] = o, o += 3;
			else ans[u] = t,t += 3;
		}
	}
	else {
		for(auto u : p[a]) {
			if(o <= n) ans[u] = o, o += 3;
			else if(z <= n) ans[u] = z, z += 3;
		}
		for(auto u : p[b]) {
			if(t <= n) ans[u] = t,t += 3;
			else if(z <= n) ans[u] = z, z += 3;
			// else ans[u] = o,o += 3;
		}
	}

	for(int i = 1;i <= n;++i) printf("%d%c", ans[i], i == n ?'\n' :' ');
}