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 JAVA):

Solution:
import java.util.*;

public class Main {
  public static int n;
  public static int[] c;
  public static ArrayList[] edge;
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    n = sc.nextInt();
    c = new int[n];
    edge = new ArrayList[n];
    for(int i = 0; i < n; i++) {
      edge[i] = new ArrayList<Integer>();
    }
    for(int i = 0; i < n - 1; i++) {
      int a = sc.nextInt() - 1;
      int b = sc.nextInt() - 1;
      edge[a].add(b);
      edge[b].add(a); 
    }
    dfs(0, -1);
    int p1 = 0;
    int p2 = 0;
    for(int i = 0; i < n; i++) {
      if(c[i] == 0) {
        p1++;
      } else {
        p2++;
      }
    }
    int[] ans = new int[n];
    int[] used = new int[n + 1];
    
    if(p1 <= (n / 3)) {
      int p = 1;
      for(int i = 0; i < n; i++) {
        if(c[i] == 0) {
          ans[i] = (3 * p);
          p++;
          used[ans[i]] = 1;
        }
      }
      ArrayList<Integer> list = new ArrayList<Integer>(); 
      for(int i = 1; i <= n; i++) {
        if(used[i] == 0) list.add(i);
      }
      int q = 0;
      for(int i = 0; i < n; i++) {
        if(c[i] == 1) {
          ans[i] = list.get(q);
          q++;
        }
      }
    } else if(p2 <= (n / 3)) {
      int p = 1;
      for(int i = 0; i < n; i++) {
        if(c[i] == 1) {
          ans[i] = (3 * p);
          p++;
          used[ans[i]] = 1;
        }
      }
      ArrayList<Integer> list = new ArrayList<Integer>(); 
      for(int i = 1; i <= n; i++) {
        if(used[i] == 0) list.add(i);
      }
      int q = 0;
      for(int i = 0; i < n; i++) {
        if(c[i] == 0) {
          ans[i] = list.get(q);
          q++;
        }
      }      
    } else {
      int p = 0;
      int q = 0;
      for(int i = 0; i < n; i++) {
        if(c[i] == 0) {
          if((3 * p + 1) <= n) {
            ans[i] = (3 * p + 1);
            p++;
          }
        } else {
        if(c[i] == 1) {
          if((3 * q + 2) <= n) {
            ans[i] = (3 * q + 2);
            q++;
          }          
        }
        }
      }
      int r = 1;
      for(int i = 0; i < n; i++) {
        if(ans[i] == 0) {
          ans[i] = (3 * r);
          r++;
        }
      }
    }
    for(int i = 0; i < n; i++) {
      System.out.println(ans[i]);
    }
  }

  public static void dfs(int v, int p) {
    for(int i = 0; i < edge[v].size(); i++) {
      int t = (int)(edge[v].get(i));
      if(t != p) {
        c[t] = 1 - c[v];
        dfs(t, v);
      }
    }    
  }
}