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 ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define IT iterator
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
#define debug puts("wzpakking")
#define y1 ysghysgsygsh
using namespace std;
const int N=200005;
vector<int> e[N];
vector<int> vec[N];
int n,vis[N],ans[N];
void color(int x,int fa,int c){
	vec[c].PB(x);
	for (auto i:e[x])
		if (i!=fa) color(i,x,c^1);
}
int main(){
	scanf("%d",&n);
	For(i,1,n-1){
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].PB(y); e[y].PB(x);
	} 
	color(1,0,0);
	if (vec[0].size()>vec[1].size())
		swap(vec[0],vec[1]);
	if (vec[0].size()<=n/3){
		int p=0;
		for (auto i:vec[0])
			ans[i]=(p+=3),vis[p]=1;
		p=1;
		for (auto i:vec[1]){
			for (;vis[p];++p);
			ans[i]=(p++);
		}
	}
	else{
		For(i,0,n/3) if (i*3+1<=n)
			ans[vec[0][i]]=i*3+1;
		For(i,0,n/3) if (i*3+2<=n)
			ans[vec[1][i]]=i*3+2;
		int p=0;
		For(i,1,n) if (!ans[i]) ans[i]=p+=3;
	}
	For(i,1,n) printf("%d ",ans[i]);
	puts("");
}
/*
按照和A的距离三染色
*/