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.BufferedOutputStream;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.util.ArrayList;
import java.util.List;

public class ER6E {

	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		
		int n = scn.nextInt();
		int m = scn.nextInt();
		int[] color = new int[n + 1];
		for (int i = 1; i <= n; ++i) {
			color[i] = scn.nextInt();
		}
		List<Integer>[] tree = new ArrayList[n + 1];
		for (int i = 0; i < n - 1; ++i) {
			int u = scn.nextInt();
			int v = scn.nextInt();
			if (tree[u] == null) {
				tree[u] = new ArrayList<>();
			}
			if (tree[v] == null) {
				tree[v] = new ArrayList<>();
			}
			tree[u].add(v);
			tree[v].add(u);
		}
		List<Integer> flat = new ArrayList<>();
		int[] entry = new int[n + 1];
		int[] exit = new int[n + 1];
		dfs(1, 0, tree, flat);
		for (int i = 0; i < flat.size(); ++i) {
			int x = flat.get(i);
			if (entry[x] == 0) {
				entry[x] = i + 1;
			} else {
				exit[x] = i + 1;
			}
		}
		int size = flat.size();
		segment = new long[4 * size];
		lazy = new long[4 * size];
		build(0, 0, size - 1, flat, color);
		for (int i = 0; i < m; ++i) {
			int t = scn.nextInt();
			if (t == 1) {
				int v = scn.nextInt();
				int c = scn.nextInt();
				int start = entry[v]-1;
				int end = exit[v]-1;
				ruupdate(0, 0, size - 1, start, end, (long) Math.pow(2, c));

			} else {
				int v = scn.nextInt();
				int start = entry[v]-1;
				int end = exit[v]-1;
				int ans = Long.bitCount(query(0, 0, size - 1, start, end));
				out.write((ans + "\n").getBytes());

			}
		}
		out.flush();
	}

	public static long query(int si, int ss, int se, int qs, int qe) {
		handlependingupdate(si, ss, se);
		if (ss >= qs && se <= qe) {
			return segment[si];
		}

		if (ss > qe || se < qs) {
			return 0;
		}

		int mid = (ss + se) / 2;
		long left = query(2 * si + 1, ss, mid, qs, qe);
		long right = query(2 * si + 2, mid + 1, se, qs, qe);
		return left | right;

	}

	public static void ruupdate(int si, int ss, int se, int us, int ue, long delta) {
		handlependingupdate(si, ss, se);

		if (ss >= us && se <= ue) {
			segment[si] = delta;
			if (ss != se) {
				lazy[2 * si + 1] = delta;
				lazy[2 * si + 2] = delta;

			}
		}

		else if (se < us || ss > ue) {
			return;
		}

		else {
			int mid = (ss + se) / 2;
			ruupdate(2 * si + 1, ss, mid, us, ue, delta);
			ruupdate(2 * si + 2, mid + 1, se, us, ue, delta);
			segment[si] = segment[2 * si + 1] | segment[2 * si + 2];
		}

	}

	public static void handlependingupdate(int si, int se, int ss) {
		if (lazy[si] != 0) {
			segment[si] = lazy[si];
			lazy[si] = 0;
			if (ss != se) {
				lazy[2 * si + 1] = segment[si];
				lazy[2 * si + 2] = segment[si];
			}
		}
	}

	public static void build(int si, int ss, int se, List<Integer> flat, int[] color) {
		if (ss == se) {
			int node = flat.get(ss);
			segment[si] = (long) Math.pow(2, color[node]);
			return;
		}

		int mid = (ss + se) / 2;
		build(2 * si + 1, ss, mid, flat, color);
		build(2 * si + 2, mid + 1, se, flat, color);
		segment[si] = segment[2 * si + 1] | segment[2 * si + 2];
	}

	static long lazy[];
	static long segment[];

	public static void dfs(int root, int p, List<Integer>[] tree, List<Integer> flat) {
		flat.add(root);

		if (tree[root] != null) {
			for (int nbr : tree[root]) {
				if (nbr != p) {
					dfs(nbr, root, tree, flat);
				}
			}
		}
		flat.add(root);
	}

	static Reader scn = new Reader();
	static OutputStream out = new BufferedOutputStream(System.out);

	static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[64]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}

}