Problem: The New Year holidays are over, but Resha doesn't want to throw away the New Year tree. He invited his best friends Kerim and Gural to help him to redecorate the New Year tree.

The New Year tree is an undirected tree with n vertices and root in the vertex 1.

You should process the queries of the two types:

  1. Change the colours of all vertices in the subtree of the vertex v to the colour c. 
  2. Find the number of different colours in the subtree of the vertex v. 

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 4·105) — the number of vertices in the tree and the number of the queries.

The second line contains n integers ci (1 ≤ ci ≤ 60) — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the vertices of the j-th edge. It is guaranteed that you are given correct undirected tree.

The last m lines contains the description of the queries. Each description starts with the integer tk (1 ≤ tk ≤ 2) — the type of the k-th query. For the queries of the first type then follows two integers vk, ck (1 ≤ vk ≤ n, 1 ≤ ck ≤ 60) — the number of the vertex whose subtree will be recoloured with the colour ck. For the queries of the second type then follows integer vk (1 ≤ vk ≤ n) — the number of the vertex for which subtree you should find the number of different colours.

Output

For each query of the second type print the integer a — the number of different colours in the subtree of the vertex given in the query.

Each of the numbers should be printed on a separate line in order of query appearing in the input.

Examples

Input

7 10
1 1 1 1 1 1 1
1 2
1 3
1 4
3 5
3 6
3 7
1 3 2
2 1
1 4 3
2 1
1 2 5
2 1
1 6 4
2 1
2 2
2 3


Output

2
3
4
5
1
2


Input

23 30
1 2 2 6 5 3 2 1 1 1 2 4 5 3 4 4 3 3 3 3 3 4 6
1 2
1 3
1 4
2 5
2 6
3 7
3 8
4 9
4 10
4 11
6 12
6 13
7 14
7 15
7 16
8 17
8 18
10 19
10 20
10 21
11 22
11 23
2 1
2 5
2 6
2 7
2 8
2 9
2 10
2 11
2 4
1 12 1
1 13 1
1 14 1
1 15 1
1 16 1
1 17 1
1 18 1
1 19 1
1 20 1
1 21 1
1 22 1
1 23 1
2 1
2 5
2 6
2 7
2 8
2 9
2 10
2 11
2 4


Output

6
1
3
3
2
1
2
3
5
5
1
2
2
1
1
1
2
3

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Reader;
import java.util.ArrayList;
import java.util.StringTokenizer;

// Educational Codeforces Round 6
// New Year Tree
public class Problem_620_E {


  public static void main(String[] args)
  {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    PrintWriter out = new PrintWriter(outputStream);
    Problem_620_E solver = new Problem_620_E();
    solver.solve(1, new InputStreamReader(inputStream), out);
    out.close();
  }

  ArrayList<Integer>[] g;
  int[] fs;
  int[] ts;
  int T = 0;
  int l;
  int r;
  int[] c;
  int ccc;
  int[] L;
  long[] seg;
  long ans;

  void dfs(int i)
  {
    if (fs[i] == 0) {
      fs[i] = ++T;
      for(int a : g[i])
        dfs(a);
      ts[i] = T;
    }
  }

  void lazyUp(int s, int e, int i){
    if(L[i] != 0) {
      seg[i] = (1L<<L[i]);
      if (s!=e){
        L[i<<1] = L[i];
        L[i<<1|1] = L[i];
      }
      L[i] = 0;
    }
  }

  void update(int s, int e, int i){
    lazyUp(s, e, i);
    if (l>e || r<s)
      return;
    if (l<=s && r>=e){
      seg[i] = (1L<<ccc);
      if (s!=e){
        L[i<<1] = ccc;
        L[i<<1|1] = ccc;
      }
      return;
    }
    int m = (s+e)>>1;
    update(s, m, i<<1);
    update(m+1, e, i<<1|1);
    seg[i] = seg[i<<1] | seg[i<<1|1];
  }

  void get(int s, int e, int i){
    if (l>e || r<s)
      return;
    lazyUp(s, e, i);
    if (l<=s && r>=e){
      ans |= seg[i];
      return;
    }
    int mid = (s+e)/2;
    get(s, mid, i<<1);
    get(mid+1, e, i<<1|1);
  }

  int bits(long x){
    int cnt = 0;
    while(x != 0) {
      cnt += (x&1L);
      x>>=1L;
    }
    return cnt;
  }
  @SuppressWarnings("unchecked")
  public void solve(int test, Reader input, PrintWriter out)
  {
    MyScanner in = new MyScanner(input);

    int n = in.nextInt();
    int m = in.nextInt();

    c = new int[n+1];
    for(int i = 1; i <= n; i++)
      c[i] = in.nextInt();

    g = new ArrayList[n+1];
    for(int i = 0; i < n+1; i++)
      g[i] = new ArrayList<Integer>();
    for(int i = 1; i < n; i++) {
      int a = in.nextInt();
      int b = in.nextInt();
      g[a].add(b);
      g[b].add(a);
    }
    ts = new int[n+1];
    fs = new int[n+1];
    seg = new long[4*n+1];
    L = new int[4*n+1];
    dfs(1);
    for(int i=1; i<=n; i++){
      l = fs[i];
      r = fs[i];
      ccc = c[i];
      update(1, T, 1);
    }
    
    for(int i = 0; i < m; i++) {
      int tp, x;
      tp = in.nextInt();
      x = in.nextInt();
      l = fs[x];
      r = ts[x];
      if (tp == 1){
          ccc = in.nextInt();
          update(1, T, 1);
      } else {
          ans = 0;
          get(1, T, 1);
          out.println(bits(ans));
      }
  }
  }

  //-----------MyScanner class for faster input----------
  public static class MyScanner {
    BufferedReader br;
    StringTokenizer st;

    public MyScanner(Reader reader) {
      br = new BufferedReader(reader);
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }

    long nextLong() {
      return Long.parseLong(next());
    }

    double nextDouble() {
      return Double.parseDouble(next());
    }

    String nextLine(){
      String str = "";
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }
      return str;
    }

  }
  //--------------------------------------------------------

}