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>
#define f(i,n) for(int i=0;i<n;i++)
#define P pair<int,int>
#define F first
#define S second
#define int long long
#define mod 1000000007
#define mmod 998244353
#define all(v) v.begin(), v.end()
using namespace std;
int mypow(int a, int b, int c) {
	if (b == 0)return 1;
	if (b % 2)return mypow(a, b - 1, c)*a%c;
	int z = mypow(a, b / 2, c);
	return z * z%c;
}
vector<int> v[200000];
int x[200000], y = 0, z = 0;
bool used[200000];
void mutu(int a, int b) {
	x[a] = b;
	if (b)z++;
	else y++;
	used[a] = true;
	f(i, v[a].size()) {
		if (!used[v[a][i]]) mutu(v[a][i], (b + 1) % 2);
	}
}
signed main() {
	int n;
	cin >> n;
	f(i, n - 1) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	mutu(0, 0);
	if (min(y, z) <= n / 3) {
		int cnt = n / 3 - min(y, z), vx = 3, zz = 1;
		f(i, n) {
			if (i)cout << " ";
			if ((int)(y > z) == x[i]) {
				cout << vx;
				vx += 3;
			}
			else if (cnt) {
				cout << vx;
				vx += 3;
				cnt--;
			}
			else {
				cout << zz;
				zz++;
				if (zz % 3 == 0)zz++;
			}
		}
		return 0;
	}
	int vy = 1, vz = 2, cnt=1;
	f(i, n) {
		if (i)cout << " ";
		if (x[i]) {
			if (vz > n) {
				cout << cnt * 3;
				cnt++;
			}
			else {
				cout << vz;
				vz += 3;
			}
		}
		else {
			if (vy > n) {
				cout << cnt * 3;
				cnt++;
			}
			else{
				cout << vy;
				vy += 3;
			}
		}
	}
}