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>
using namespace std;
int col[200005],p[200005],cnt0,cnt1;
vector<int> e[200005];
void dfs(int u,int fa)
{
 col[u]=!col[fa];
 if(col[u])cnt1++;
 else cnt0++;
 for(auto v:e[u])
  if(v!=fa)dfs(v,u);
}
int main()
{
 int n;
 cin>>n;
 for(int i=1;i<n;i++)
 {
  int u,v;
  cin>>u>>v;
  e[u].push_back(v);
  e[v].push_back(u);
 }
 dfs(1,1);
 int p1=1,p2=2,p3=3;
 if(cnt0>n/3&&cnt1>n/3)
 {
  for(int i=1;i<=n;i++)
   if(col[i]==0)
   {
    if(p1<=n)p[i]=p1,p1+=3;
    else p[i]=p3,p3+=3;
   }
   else
   {
    if(p2<=n)p[i]=p2,p2+=3;
    else p[i]=p3,p3+=3;
   }
 }
 else if(cnt1<=n/3)
 {
  for(int i=1;i<=n;i++)
   if(col[i]==0)
   {
    if(p1<=n)p[i]=p1,p1+=3;
    else if(p2<=n)p[i]=p2,p2+=3;
    else p[i]=p3,p3+=3;
   }
   else
    p[i]=p3,p3+=3;
 }
 else if(cnt0<=n/3)
 {
  for(int i=1;i<=n;i++)
   if(col[i]==1)
   {
    if(p1<=n)p[i]=p1,p1+=3;
    else if(p2<=n)p[i]=p2,p2+=3;
    else p[i]=p3,p3+=3;
   }
   else
    p[i]=p3,p3+=3;
 }
 for(int i=1;i<=n;i++)
  cout<<p[i]<<' ';
 cout<<endl;
 return 0;
}