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:

public class Main {

  private static void solve() {
    int n = ni();
    int[] from = new int[n - 1];
    int[] to = new int[n - 1];

    for (int i = 0; i < n - 1; i++) {
      from[i] = ni() - 1;
      to[i] = ni() - 1;
    }
    int[][] g = packU(n, from, to);

    int[] a = new int[n];
    dfs(0, -1, g, a);

    int[] cnt = new int[2];
    for (int v : a) {
      cnt[v]++;
    }
    if (cnt[0] > cnt[1]) {
      for (int i = 0; i < n; i++) {
        a[i] ^= 1;
      }
    }

    int[] ret = new int[n];
    int[] p = {3, 1, 2};
    int k;
    if (cnt[0] > n / 3 && cnt[1] > n / 3) {
      for (int i = 0; i < n; i++) {
        if (a[i] == 0) {
          k = p[1] <= n ? 1 : 0;
        } else {
          k = p[2] <= n ? 2 : 0;
        }
        ret[i] = p[k];
        p[k] += 3;
      }
    } else {
      for (int i = 0; i < n; i++) {
        if (a[i] == 0) {
          k = 0;
        } else {
          k = p[1] <= n ? 1 : p[2] <= n ? 2 : 0;
        }
        ret[i] = p[k];
        p[k] += 3;
      }
    }

    StringBuilder sb = new StringBuilder();
    for (int v : ret) {
      sb.append(v + " ");
    }
    System.out.println(sb);
  }

  private static final void dfs(int n, int pre, int[][] g, int[] a) {
    for (int nex : g[n]) {
      if (nex == pre)
        continue;

      a[nex] = a[n] ^ 1;
      dfs(nex, n, g, a);
    }
  }


  static int[][] packU(int n, int[] from, int[] to) {
    int[][] g = new int[n][];
    int[] p = new int[n];
    for (int f : from)
      p[f]++;
    for (int t : to)
      p[t]++;
    for (int i = 0; i < n; i++)
      g[i] = new int[p[i]];
    for (int i = 0; i < from.length; i++) {
      g[from[i]][--p[from[i]]] = to[i];
      g[to[i]][--p[to[i]]] = from[i];
    }
    return g;
  }

  public static void main(String[] args) {
    new Thread(null, new Runnable() {
      @Override
      public void run() {
        long start = System.currentTimeMillis();
        String debug = args.length > 0 ? args[0] : null;
        if (debug != null) {
          try {
            is = java.nio.file.Files.newInputStream(java.nio.file.Paths.get(debug));
          } catch (Exception e) {
            throw new RuntimeException(e);
          }
        }
        reader = new java.io.BufferedReader(new java.io.InputStreamReader(is), 32768);
        solve();
        out.flush();
        tr((System.currentTimeMillis() - start) + "ms");
      }
    }, "", 64000000).start();
  }

  private static java.io.InputStream is = System.in;
  private static java.io.PrintWriter out = new java.io.PrintWriter(System.out);
  private static java.util.StringTokenizer tokenizer = null;
  private static java.io.BufferedReader reader;

  public static String next() {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      try {
        tokenizer = new java.util.StringTokenizer(reader.readLine());
      } catch (Exception e) {
        throw new RuntimeException(e);
      }
    }
    return tokenizer.nextToken();
  }

  private static double nd() {
    return Double.parseDouble(next());
  }

  private static long nl() {
    return Long.parseLong(next());
  }

  private static int[] na(int n) {
    int[] a = new int[n];
    for (int i = 0; i < n; i++)
      a[i] = ni();
    return a;
  }

  private static char[] ns() {
    return next().toCharArray();
  }

  private static long[] nal(int n) {
    long[] a = new long[n];
    for (int i = 0; i < n; i++)
      a[i] = nl();
    return a;
  }

  private static int[][] ntable(int n, int m) {
    int[][] table = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[i][j] = ni();
      }
    }
    return table;
  }

  private static int[][] nlist(int n, int m) {
    int[][] table = new int[m][n];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[j][i] = ni();
      }
    }
    return table;
  }

  private static int ni() {
    return Integer.parseInt(next());
  }

  private static void tr(Object... o) {
    if (is != System.in)
      System.out.println(java.util.Arrays.deepToString(o));
  }
}