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 pb push_back
const int maxi = 1e6+10;
vector<int> v[maxi], l[2];
int n;
int rez[maxi], ob[maxi];

void dfs(int x, int pred, int lev)
{
    l[lev&1].pb(x);

    for (auto i:v[x])
        if (i!=pred) dfs(i, x, lev+1);
}
int main()
{
   cin>>n;

  for (int i=1;i<n;i++)
  {
      int x, y;
      scanf("%d%d",&x,&y);
      v[x].pb(y);
      v[y].pb(x);
  }

    dfs(1, 0, 0);

    if (l[0].size()>l[1].size()) swap(l[0], l[1]);

    int cur = 0;

    if (l[0].size()<=n/3)
    {
       for (int i = 3;i<=n && l[0].size()>cur;i+=3)
       {
           rez[l[0][cur]] = i;
           ob[i] = 1;
           cur++;
       }
    }
  else
  {
      for (int i = 1;i<=n && l[0].size()>cur;i+=3)
      {
          rez[l[0][cur]] = i;
          ob[i] = 1;
          cur++;
      }

      for (int i = 3;i<=n && l[0].size()>cur;i+=3)
      {
          rez[l[0][cur]] = i;
          ob[i] = 1;
          cur++;
      }
  }

      cur = 0;
      for (int i =1;i<=n;i++)
        if (!ob[i])
      {
          rez[l[1][cur]] = i;
          cur++;
      }

  for (int i=1;i<=n;i++)
    printf("%d ", rez[i]);
}