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 long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
struct edge
{
  ll to;
  ll cost;
  edge() {}
  edge(ll to, ll cost) : to(to), cost(cost){};
};
int main()
{
ll n;
cin >> n ;
vector<vector<edge>> g(n);
rep(i, n-1)
{
  ll a, b;
  cin >> a >> b ;
  ll c=0;
  g[a - 1].emplace_back(b - 1, c);
  g[b - 1].emplace_back(a - 1, c); //無向辺
}
ll a=1,b=2,c=3;
ll ev=1,od=0;
vector<ll> par(n);
vector<ll> res(n);
vector<bool> see(n);
queue<ll> q;
par[0]=-1;
see[0]=1;
res[0]=0;
q.push(0);
while(!q.empty()){
  auto p=q.front();
  see[p]=1;
  q.pop();
  for(auto r:g[p]){
    if(see[r.to]!=0)continue;
    par[r.to]=p;
    res[r.to]=res[p]+1;
    if(res[r.to]%2)od++;
    else ev++;
    q.push(r.to);
  }
}
vector<ll> ans(n,-1);
if(od>n/3&&ev>n/3){
  rep(i,n){
    if(res[i]%2){
      if(a>n) {ans[i]=c;c+=3;}
      else{ans[i]=a;a+=3;}
    }
    else{
      if(b>n) {ans[i]=c;c+=3;}
      else{ans[i]=b;b+=3;}
    }
  }
}
else if(od>n/3)
{
  rep(i,n){
    if(res[i]%2==0){ans[i]=c;c+=3;}
  }
  rep(i,n){
    if(res[i]%2==1&&c<=n){ans[i]=c;c+=3;}
    else if(res[i]%2==1&&a<=n){ans[i]=a;a+=3;}
    else if(res[i]%2==1){ans[i]=b;b+=3;}
  }
}
else
{
  rep(i,n){
    if(res[i]%2==1){ans[i]=c;c+=3;}
  }
  rep(i,n){
    if(res[i]%2==0&&c<=n){ans[i]=c;c+=3;}
    else if(res[i]%2==0&&a<=n){ans[i]=a;a+=3;}
    else if(res[i]%2==0){ans[i]=b;b+=3;}
  }
}

for(auto r : ans)cout << r << ' ';
cout << endl;
return 0;
}