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<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(int i=0;i<int(n);i++)
#define N 200020
vector<int>e[N];
int cc[2];
int c[N];
vector<int>ca[2];
void dfs(int k, int nc) {
	c[k] = nc;
	ca[nc].push_back(k);
	cc[nc]++;
	f(i, e[k].size()) {
		if (c[e[k][i]] == -1)dfs(e[k][i], ((nc + 1) % 2));
	}
	return;
}


int main() {
	int a[N];
	f(i, N)a[i] = 0;
	int n, k;
	int x, y, z;
	long long s, ans;
	bool v = true;
	ans = 0;
	scanf("%d", &n);
	f(i, n-1) {
		scanf("%d %d", &x, &y);
		e[x - 1].push_back(y - 1);
		e[y - 1].push_back(x - 1);
	}
	f(i, n) {
		c[i] = -1;
	}
	cc[0] = 0;
	cc[1] = 0;
	dfs(0, 0);
	x = (n + 2) / 3;
	if (cc[0] >= x) {
		cc[0] -= x;
		f(i, x)a[ca[0][i]] = (3 * i) + 1;
	}
	else {
		cc[1] -= x;
		f(i, x)a[ca[1][i]] = (3 * i) + 1;
	}
	y = (n + 1) / 3;
	if (cc[0] < cc[1]) {
		z = ca[1].size();
		f(i, y) {
			a[ca[1][z - i - 1]] = (3 * i) + 2;
		}
	}
	else {
		z = ca[0].size();
		f(i, y) {
			a[ca[0][z - i - 1]] = (3 * i) + 2;
		}
	}
	x = 3;
	f(i, n) {
		if (a[i] == 0) {
			a[i] = x;
			x += 3;
		}
	}
	f(i, n) {
		printf("%d", a[i]);
		if (i < (n - 1))printf(" ");
		else printf("\n");
	}


	return 0;
}