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>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define rrep(i,a,b) for(int i=(a);i>=(b);--i)
#define PB push_back
#define ar2 array<int,2> 
#define vs vector<string>
typedef long long LL;
const LL P = 1e9+7;
const int N = 2e5+5;
mt19937 rng(time(0));

vector<int> g[N];
int n, x, y, a[N], r[4];

void dfs(int x, int fa, int num){
	a[x] = num;
	for(auto u:g[x])
		if(u!=fa) dfs(u, x, 1-num);
}

int main() {
	scanf("%d",&n);
	rep(i,1,n-1){
		scanf("%d%d",&x,&y);
		g[x].PB(y), g[y].PB(x);
	}
	dfs(1,0,1);
	int cnt = 0;
	rep(i,1,n) cnt += a[i];
	if(cnt*2>n){
		rep(i,1,n) a[i] = 1-a[i];
		cnt = n - cnt;
	}
	
	if(cnt*3<=n){
		rep(i,1,n) if(a[i]==1) a[i] = 3;
		r[1] = (n+2)/3, r[2] = (n+1)/3, r[3] = n/3-cnt;
		rep(i,1,n) if(a[i]==0){
			rep(j,1,3) if(r[j]){
				r[j]--, a[i] = j;
				break;
			}
		}
	}else{
		rep(i,1,n) if(a[i]==0) a[i] = 1; else a[i] = 2;
		r[1] = n - cnt - (n+2)/3, r[2] = cnt - (n+1)/3;
		rep(i,1,n){
			int x = a[i];
			if(!r[x]) continue;
			r[x]--, a[i] = 3;
		}
	}
	rep(i,1,3) r[i] = i;
	rep(i,1,n) x = r[a[i]], r[a[i]] += 3, a[i] = x;
	rep(i,1,n) printf("%d ",a[i]);
	return 0;
}