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:
#pragma GCC optimize ("Ofast")
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i=(a); i<(int)(b); i++)
#define FORD(i, a, b) for (int i=a; i>(int)(b); i--)
#define PPC(x) __builtin_popcount(x)
#define MSB(x) (31 - __builtin_clz(x))
#define pb push_back
#define ALL(x) (x).begin(), (x).end()
#define ithBit(m, i) ((m) >> (i) & 1)
#define ft first
#define sd second
#ifdef DEBUG
#include "debug.h"
#else
#define dbg(...) 0
#endif
using namespace std;

const int maxN = 1 << 19, mod = 1000000007;

vector <int> G[maxN], A, B, r[3];
int ile[3], p[maxN];

void dfs(int v, int fat, int d)
{
	d % 2 ? B.pb(v) : A.pb(v);
	for (int u : G[v])
		if (u != fat)
			dfs(u, v, d+1);
}

void mv(vector <int>& v, int k)
{
	r[k].pb(v.back());
	v.pop_back();
	ile[k]--;
}

void solve()
{
	int n;
	scanf ("%d", &n);
	FOR(i, 1, n)
	{
		int a, b;
		scanf ("%d%d", &a, &b);
		G[a].pb(b);
		G[b].pb(a);
	}
	
	dfs(1, 0, 0);
	
	ile[0] = n / 3;
	ile[1] = (n+2) / 3;
	ile[2] = n/3 + (n%3 == 2);
	
	if (A.size() > B.size())
		swap(A, B);
	
	while (!B.empty() and ile[2])
		mv(B, 2);
		
	vector <int>& C = A.size() > B.size() ? A : B;
	while (ile[1])
		mv(C, 1);
		
	while (ile[0])
	{
		if (A.empty())
			mv(B, 0);
		else
			mv(A, 0);
	}
	
	FOR(k, 1, 4)
	{
		int i = k;
		for (int v : r[k%3])
			p[v] = i, i+=3;
	}
	
	FOR(i, 1, n+1)
		printf("%d ", p[i]);
	printf("\n");	
} 

 
int main()
{
    int t;
    t=1;//scanf ("%d", &t);
    while (t--)
    	solve();
    return 0;    
}