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 <iostream>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
typedef long long int ll;

vector<int> g[200200];
int color[200200];
int ans[200200];
int cnt=0;

void dfs(int s){
	if(color[s]==0)cnt++;
	for(int t:g[s]){
		if(color[t]!=-1)continue;
		color[t]=color[s]^1;
		dfs(t);
	}
}

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n; cin >> n;
	for(int i=1;i<n;i++){
		int x,y; cin >> x >> y;
		x--; y--;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	for(int i=0;i<n;i++){
		color[i]=-1;
	}
	color[0]=0;
	dfs(0);
	if(cnt<=n/3){
		for(int i=0,j=1;i<n;i++){
			if(color[i]==0){
				ans[i]=3*j;
				j++;
			}
		}
	}
	else if((n-cnt)<=n/3){
		for(int i=0,j=1;i<n;i++){
			if(color[i]==1){
				ans[i]=3*j;
				j++;
			}
		}
	}
	else{
		for(int i=0,j=1,k=1;i<n;i++){
			if(color[i]==0&&3*j-2<=n){
				ans[i]=3*j-2;
				j++;
			}
			if(color[i]==1&&3*k-1<=n){
				ans[i]=3*k-1;
				k++;
			}
		}
	}
	set<int> st;
	for(int i=1;i<=n;i++){
		st.insert(i);
	}
	for(int i=0;i<n;i++){
		if(ans[i]){
			st.erase(ans[i]);
		}
	}
	for(int i=0;i<n;i++){
		if(ans[i]==0){
			ans[i]=*st.begin();
			st.erase(st.begin());
		}
	}
	for(int i=0;i<n;i++){
		printf("%d\n",ans[i]);
	}
}