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>
using namespace std;
const int N=200000+10;
int n,a[N],b[N],p[N],dep[N];
vector<int> g[N];
void dfs(int u,int p){
	dep[u]=dep[p]+1;
	for(auto v:g[u]){
		if(v==p)continue;
		dfs(v,u);
	}
}
/*
3 3 2

min(odd, even): <= 2
min(odd, even) >= 3
*/
int main() {
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		scanf("%d%d",&a[i],&b[i]);
		g[a[i]].push_back(b[i]);
		g[b[i]].push_back(a[i]);
	}
	dfs(1,1);
	int d1=n/3,d2=n/3,d3=n/3;
	if(n%3==1) d1++;
	if(n%3==2) d1++,d2++;
	int c1=0,c2=0;
	for(int i=1;i<=n;i++)if(dep[i]%2==0)c1++;else c2++;

	int x=1,y=2,z=3;
	if(min(c1,c2)<=d3) {
		int bu = c1<c2 ? 0 : 1;
		for(int i=1;i<=n;i++){
			if(dep[i]%2==bu){
				p[i]=z; z+=3;
			} else {
				if(x<=n) p[i]=x,x+=3;
				else if(y<=n) p[i]=y,y+=3;
				else p[i]=z,z+=3;
			}
		}
	} else {
		for(int i=1;i<=n;i++){
			if(dep[i]%2==0){
				if(x<=n){
					p[i]=x; x+=3;
				}else{
					p[i]=z; z+=3;
				}
			} else {
				if(y<=n){
					p[i]=y; y+=3;
				}else{
					p[i]=z; z+=3;
				}
			}
		}
	}
	for(int i=1;i<=n;i++)printf("%d ", p[i]);
	
}