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 rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair
 
using namespace std;
 
const LL MOD=1e9+7;

int n,x,y,ans[200010];
vector <int> g[200010],v[2],num[3],allpos,allval;

void dfs(int pos,int no,int d)
{
	v[d].pb(pos);
	rep(i,g[pos].size()) if(g[pos][i]!=no) dfs(g[pos][i],pos,d^1);
}

int main()
{
	cin>>n;
	rep(i,n-1)
	{
		scanf("%d%d",&x,&y);
		g[x].pb(y);
		g[y].pb(x);
	}
	dfs(1,0,0);
	repn(i,n) num[i%3].pb(i);
	if(v[0].size()<v[1].size()) swap(v[0],v[1]);
	rep(i,v[0].size()) allpos.pb(v[0][i]);
	rep(i,v[1].size()) allpos.pb(v[1][i]);
	if(v[1].size()>=num[2].size())
	{
		rep(i,num[1].size()) allval.pb(num[1][i]);
		rep(i,num[0].size()) allval.pb(num[0][i]);
		rep(i,num[2].size()) allval.pb(num[2][i]);
	}
	else
	{
		rep(i,num[1].size()) allval.pb(num[1][i]);
		rep(i,num[2].size()) allval.pb(num[2][i]);
		rep(i,num[0].size()) allval.pb(num[0][i]);
	}
	rep(i,n) ans[allpos[i]]=allval[i];
	repn(i,n) cout<<ans[i]<<' ';
	cout<<endl;
	return 0;
}