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,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se seoncd
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
 
template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=200005;
int col[N],ans[N],n,u,v;
vi e[N],x,y,a,b,c;

void dfs(int u,int fa){
	if(col[u]) y.pb(u);
	else x.pb(u);
	for(auto v:e[u]){
		if(v==fa) continue;
		col[v]=col[u]^1;
		dfs(v,u);
	}
}

void doit(vi &x,vi &c){
	while(x.size()&&c.size()){
		ans[x.back()]=c.back();
		x.pop_back();
		c.pop_back();
	}
}

int main(){
	read(n);
	rep(i,1,n-1){
		read(u),read(v);
		e[u].pb(v),e[v].pb(u);
	}
	dfs(1,0);
	if(x.size()>y.size()) swap(x,y);
	rep(i,1,n){
		if(i%3==1) a.pb(i);
		else if(i%3==2) b.pb(i);
		else c.pb(i);
	}
	if(x.size()<=c.size()){
		doit(x,c);
		doit(y,c);
		doit(y,a);
		doit(y,b);
	}
	else{
		doit(x,a);
		doit(y,b);
		doit(x,c);
		doit(y,c);
	}
	rep(i,1,n) printf("%d ",ans[i]);
	return 0;
}