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 ll long long
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB emplace_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
#define lwb lower_bound
#define upb upper_bound
using namespace std;
int co[200005],n,sum[2];
V<int> v[200005];
void dfs(int p,int fa,int color){
	co[p]=color;
	sum[co[p]]++;
	for(auto u:v[p]){
		if(u==fa)continue;
		dfs(u,p,color^1);
	}
}
int ans[200005];
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;
	int x,y;
	rep(i,1,n){
		cin>>x>>y;
		v[x].PB(y);
		v[y].PB(x);
	}
	dfs(1,-1,0);
	int l[4];
	l[1]=n/3+(n%3>0);
	l[2]=n/3+(n%3>1);
	l[3]=n/3;
	int t[2]={};
	if(l[1]<=sum[0]&&l[2]<=sum[1]){
		t[0]=1;t[1]=2;
	}else if(l[1]<=sum[1]&&l[2]<=sum[0]){
		t[1]=1;t[0]=2;
	}
	if(t[0]){
		FOR(i,1,n)ans[i]=t[co[i]],l[t[co[i]]]--;
		FOR(i,1,n){
			if(co[i]){
				if(l[t[1]]<0){
					l[t[1]]++;
					ans[i]=3;
				}
			}else{
				if(l[t[0]]<0){
					l[t[0]]++;
					ans[i]=3;
				}
			}
		}
	}else{
		int lc;
		if(l[3]>=sum[0]){
			lc=0;
		}else lc=1;
		FOR(i,1,n){
			if(co[i]==lc)ans[i]=3;
			else if(l[1]){
				l[1]--;ans[i]=1;
			}else if(l[2]){
				l[2]--;ans[i]=2;
			}else ans[i]=3; 
		}
	}
	int st[4]={};
	FOR(i,1,n){
		cout<<st[ans[i]]*3+ans[i]<<' ';
		st[ans[i]]++;
	}
	RE 0;
}