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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;


public class E {

	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
	static ArrayList<Integer>[]ages;
	static int[]order, tin, tout;
	static int size, timer, n;
	static boolean[]used;
	static long[]tree;
	static int[]t_color;
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		n = nextInt();
		int m = nextInt();
		order = new int[n+1];
		used = new boolean[n+1];
		tin = new int[n+1];
		tout = new int[n+1];
		int[]c = new int[n+1];
		for (int i = 1; i <= n; i++) {
			c[i] = nextInt();
		}
		ages = new ArrayList[n+1];
		for (int i = 1; i <= n; i++) {
			ages[i] = new ArrayList<>();
		}
		for (int i = 1; i < n; i++) {
			int v1 = nextInt();
			int v2 = nextInt();
			ages[v1].add(v2);
			ages[v2].add(v1);
		}
		dfs(1);
		int[]first = new int[n+1];
		for (int i = 1; i <= n; i++) {
			first[order[i]] = i;
		}
		tree = new long[4*n];
		t_color = new int[4*n];
		for (int i = 1; i <= n; i++) {
			update(1, 1, n, first[i], first[i], c[i]);
		}
		while (m-->0) {
			int t = nextInt();
			int v = nextInt();
			int left = first[v], right = n + 1;
			while (right-left > 1) {
				int mid = (left+right) >> 1;
				if (tin[order[mid]] > tout[v])
					right = mid;
				else
					left = mid;
			}
			if (t==1) {
				int cx = nextInt();
				update(1, 1, n, first[v], left, cx);
			}
			else {
				long col = get(1, 1, n, first[v], left);
				pw.println(Long.bitCount(col));
			}
		}
		pw.close();
	}
	private static long get(int v, int tl, int tr, int L, int R) {
		if (L > R)
			return 0;
		if (tl==L && tr==R)
			return tree[v];
		push(v);
		int tm = (tl+tr) >> 1;
		return (get(2*v, tl, tm, L, Math.min(tm, R)) | get(2*v+1, tm+1, tr, Math.max(L, tm+1), R));
	}
	private static void update(int v, int tl, int tr, int L, int R, int c) {
		if (L > R)
			return;
		if (tl==L && tr==R) {
			t_color[v] = c;
			tree[v] = (1l << (c-1));
			return;
		}
		push(v);
		int tm = (tl+tr) >> 1;
		update(2*v, tl, tm, L, Math.min(tm, R), c);
		update(2*v+1, tm+1, tr, Math.max(tm+1, L), R, c);
		tree[v] = (tree[2*v] | tree[2*v+1]);
	}
	private static void push(int v) {
		if (t_color[v] != 0) {
			if (2*v < 4*n) {
				t_color[2*v] = t_color[v];
				tree[2*v] = (1l << (t_color[v]-1));
			}
			if (2*v+1 < 4*n) {
				t_color[2*v+1] = t_color[v];
				tree[2*v+1] = (1l << (t_color[v]-1));
			}
			t_color[v] = 0;
		}
	}
	private static void dfs(int v) {
		order[++size] = v;
		used[v] = true;
		tin[v] = timer++;
		for (int to : ages[v]) {
			if (!used[to])
				dfs(to);
		}
		tout[v] = timer++;
	}
	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	private static String next() throws IOException {
		while (st==null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}