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:
//Solution by Tima
#include <bits/stdc++.h>
 
#define f first
#define s second
#define ll long long
#define ull unsigned long long
#define mp make_pair
#define pb push_back
#define vi vector <int>
#define ld long double
#define pii pair<int, int>
#define y1 sda
 
using namespace std;    
const int N = int(3e5) + 12, mod = int(1e9)  + 7; 

int c[3];
int n,p[N];
vector <int> g[N], rb[N];

void dfs(int v,int col, int p){
	rb[col].pb(v);
	for(int to : g[v]){
		if(to == p) continue;
		dfs(to, 3 - col, v);
	}
}

int main () {
	scanf("%d", &n);
	for(int i = 1,u,v; i < n; i++){
		scanf("%d%d",&u, &v);
		g[u].pb(v);
		g[v].pb(u);
	}
	c[0] = 3, c[1] = 1, c[2] = 2;
	dfs(1,1,-1);
	if(rb[1].size() > rb[2].size()) swap(rb[1], rb[2]);
	if(rb[1].size() <= int(n / 3)){
		for(int i : rb[1]){
		    p[i] = c[0];
		    c[0] += 3;
		}
		for(int i : rb[2]){
			if(c[0] <= n){
				p[i] = c[0];
			    c[0] += 3;
			}
			else if(c[1] <= n){
				p[i] = c[1];
			    c[1] += 3;
			}
			else if(c[2] <= n){
				p[i] = c[2];
			    c[2] += 3;
			}
		}
	}
	else{
		for(int i : rb[1]){
			if(c[1] <= n){
				p[i] = c[1];
				c[1] += 3;
			}
			else{
				p[i] = c[0];
				c[0] += 3;
			}
		}
		for(int i : rb[2]){
			if(c[2] <= n){
				p[i] = c[2];
				c[2] += 3;
			}
			else{
				p[i] = c[0];
				c[0] += 3;
			}
		}
	}
	for(int i = 1; i <= n; i++) printf("%d ", p[i]);

return 0;
}