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;
typedef pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second

const int mx=200010;
const ll mod=1e9+7;

vector<int> v[mx];
vector<int> g(mx,-1);
  
void dfs(int n, int p, int dep){
  g[n] = dep%2;
  for(auto to:v[n])if(to!=p){ dfs(to,n,dep+1); }
}

int main() {
  int n;
  cin >> n;
  rep(i,n-1){
    int a,b; cin >> a >> b; a--; b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(0,-1,0);

  int nb = count(g.begin(),g.end(),0);
  int nw = count(g.begin(),g.end(),1);
  if(nb>nw){ swap(nb,nw); rep(i,n){g[i]=1-g[i];} }
  
  vector<int> ans(n);
  if(nb<=n/3){
    int bcnt=3,wcnt=1;
    rep(i,n){
      if(g[i]==0){ 
        ans[i]=bcnt; bcnt+=3; 
      } else {
        ans[i]=wcnt; wcnt+=(wcnt%3==1?1:2);
      }
    }
    rep(i,n)if(ans[i]>n){ ans[i]=bcnt; bcnt+=3; }

  } else {
    int bcnt=1,wcnt=2;
    rep(i,n){
      if(g[i]==0){ 
        ans[i]=bcnt; bcnt+=3; 
      } else {
        ans[i]=wcnt; wcnt+=3;
      }
    }
    int cnt=3;
    rep(i,n)if(ans[i]>n){ ans[i]=cnt; cnt+=3; }
  }
  rep(i,n){ cout << ans[i] << " "; }
  cout << endl;
  return 0;
}