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.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class E {
	static int t, tin[], tout[];
	static ArrayList<Integer> adj[];
	static int[] color;
	static boolean[] vis;
	static long[] arr;

	static void dfs(int u) {
		vis[u] = true;
		tin[u] = ++t;
		arr[t] = 1L << color[u];
		for (int v : adj[u])
			if (!vis[v]) {
				dfs(v);
			}
		tout[u] = t;
	}

	public static void main(String[] args) throws Throwable {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int q = sc.nextInt();
		color = new int[n];
		for (int i = 0; i < n; i++)
			color[i] = sc.nextInt();
		int N = 1;
		while (N < n)
			N <<= 1;
		adj = new ArrayList[n];
		for (int i = 0; i < n; i++)
			adj[i] = new ArrayList<>();
		for (int i = 0; i < n - 1; i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			adj[u].add(v);
			adj[v].add(u);
		}
		t = 0;
		arr = new long[N + 1];
		tin = new int[n + 1];
		tout = new int[n + 1];
		vis = new boolean[n];
		dfs(0);
		SegmentTree st = new SegmentTree(arr);
		PrintWriter pw = new PrintWriter(System.out);
		while (q-- > 0) {
			int type = sc.nextInt();
			if (type == 1) {
				int u = sc.nextInt() - 1;
				int c = sc.nextInt();
				st.update_range(tin[u], tout[u], 1L << c);
			}
			else {
				int u = sc.nextInt() - 1;
				pw.println(Long.bitCount(st.query(tin[u], tout[u])));
			}
		}
		pw.flush();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(new File(s)));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

		public boolean ready() throws IOException {
			return br.ready();
		}
	}

	static class SegmentTree { // 1-based DS, OOP
		//1-indexed  1-indexed  1-indexed  1-indexed  1-indexed  1-indexed  1-indexed  1-indexed  1-indexed  1-indexed  1-indexed  1-indexed  
		int N;
		long[] array, sTree, lazy;

		SegmentTree(long[] in)
		{
			array = in;
			N = in.length - 1;
			sTree = new long[N << 1];
			lazy = new long[N << 1];
			build(1, 1, N);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				build(node << 1, b, (b + e) / 2);
				build((node << 1) + 1, (b + e) / 2 + 1, e);
				sTree[node] = sTree[node << 1] | sTree[(node << 1) + 1];
			}
		}

		void update_range(int i, int j, long val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, long val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] = val;
				lazy[node] = val;
			}
			else {
				propagate(node, b, e);
				update_range(node << 1, b, (b + e) / 2, i, j, val);
				update_range((node << 1) + 1, (b + e) / 2 + 1, e, i, j, val);
				sTree[node] = sTree[node << 1] | sTree[(node << 1) + 1];
			}
		}

		void propagate(int node, int b, int e) {
			if (lazy[node] != 0) {
				lazy[node << 1] = lazy[node];
				lazy[(node << 1) + 1] = lazy[node];
				sTree[node << 1] = lazy[node];
				sTree[(node << 1) + 1] = lazy[node];
			}
			lazy[node] = 0;
		}

		long query(int i, int j) {
			return query(1, 1, N, i, j);
		}

		long query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return sTree[node];
			propagate(node, b, e);
			long q1 = query(node << 1, b, (b + e) / 2, i, j);
			long q2 = query((node << 1) + 1, (b + e) / 2 + 1, e, i, j);
			return q1 | q2;

		}
	}
}