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<cstdio>
#include<algorithm>
#include<vector>
#define N_ 201000
using namespace std;
vector<int>E[N_], G[3], V[2];
int T[N_], n, Col[N_], chk[N_], Res[N_];
void DFS(int a, int pp) {
	Col[a] = !Col[pp];
	V[Col[a]].push_back(a);
	for(auto &x : E[a]) {
		if (x == pp)continue;
		DFS(x, a);
	}
}
int main() {
	int i, a, b;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		G[i % 3].push_back(i);
	}
	for (i = 1; i < n; i++) {
		scanf("%d%d", &a, &b);
		E[a].push_back(b);
		E[b].push_back(a);
	}
	DFS(1, 0);
	int pv = 0;
	if (V[pv].size() > V[!pv].size())pv = !pv;
	if (V[pv].size() <= G[0].size()) {
		int pp = 0;
		for (auto &x : V[pv]) {
			Res[x] = G[0][pp];
			chk[G[0][pp]] = 1;
			pp++;
		}
	}
	else {
		int pp = 0, pp2 = 0;
		for (auto &x : V[pv]) {
			if (pp < G[1].size()) {
				Res[x] = G[1][pp];
				chk[G[1][pp]] = 1;
				pp++;
			}
			else {
				Res[x] = G[0][pp2];
				chk[G[0][pp2]] = 1;
				pp2++;
			}
		}
	}
	vector<int>TP;
	for (i = 1; i <= n; i++)if (!chk[i])TP.push_back(i);
	for (i = 0; i < V[!pv].size(); i++) {
		Res[V[!pv][i]] = TP[i];
	}
	for (i = 1; i <= n; i++)printf("%d ", Res[i]);
}