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.util.*;
import java.io.*;
public class NewYearTree
{


	/************************ SOLUTION STARTS HERE ***********************/

	static ArrayList<Integer>[] adj;
	static int L[] , R[] , time = 1, newToOld[];

	static void dfs(int u , int par) {
		L[u] = time;
		newToOld[time] = u;
		time++;
		for(int v : adj[u])
			if(v != par)
				dfs(v, u);

		R[u] = time - 1;
	}

	static class SegmentTreeLazyProp {

		long tree[];
		int lazy[];
		int len;
		int size;
		SegmentTreeLazyProp(int arr[]) { // one based index
			len = arr.length - 1;
			size = 1 << (32 - Integer.numberOfLeadingZeros(len - 1) + 1);  // ceil(log(len)) + 1
			tree = new long[size];
			lazy = new  int[size];
			build(1, 1, len, arr);
		}

		long build(int n , int nl , int nr , int arr[]) {
			if(nl == nr) 
				return tree[n] = 1L << arr[newToOld[nl]];
			else {
				int mid = (nl + nr) / 2;
				return tree[n] = build(2*n, nl, mid, arr) | build(2*n + 1, mid + 1, nr, arr);
			}
		}

		boolean isInternal(int n) {
			return 2*n < size && 2*n + 1 < size;
		}

		void upd(int n) {
			if(lazy[n] > 0) {
				if(isInternal(n)) {
					lazy[2*n] = lazy[n];
					lazy[2*n + 1] = lazy[n];
				}
				tree[n] = 1L << lazy[n];
				lazy[n] = 0;
			}
		}

		long update(int n , int nl , int nr , int L ,int R, int c) {

			if(nl == L && nr == R) {
				if(isInternal(n)) {
					lazy[2*n] = c;
					lazy[2*n + 1] = c; 
				}
				lazy[n] = 0;
				return tree[n] = 1L << c;
			}
			else {
				upd(n);
				if(isInternal(n)) {
					upd(2*n);
					upd(2*n + 1);
				}

				long left = tree[2*n];
				long right = tree[2*n + 1];
				int mid = (nl + nr) / 2;

				if(R <= mid)
					left = update(2*n, nl, mid, L, R, c);
				else if(L > mid)
					right = update(2*n + 1, mid + 1, nr, L, R, c);
				else {
					left = update(2*n, nl, mid, L, mid, c);
					right = update(2*n + 1, mid + 1, nr, mid + 1, R, c);
				}

				return tree[n] = left | right;

			}
		}

		void update(int v , int c) {
			update(1, 1, len, L[v], R[v], c);
		}

		long query(int n , int nl , int nr , int L , int R) {
			upd(n);
			if(nl == L && nr == R)
				return tree[n];
			else {
				int mid = (nl + nr) / 2;
				if(R <= mid)
					return query(2*n, nl, mid, L, R);
				else if(L > mid)
					return query(2*n + 1, mid + 1, nr, L, R);
				else
					return query(2*n, nl, mid, L, mid) | query(2*n + 1, mid + 1, nr, mid + 1, R);

			}
		}

		int query(int v) {
			return Long.bitCount(query(1, 1, len, L[v], R[v]));
		}

	}

	@SuppressWarnings("unchecked")
	private static void solve(FastScanner s1, PrintWriter out){

		int V = s1.nextInt();
		int E = V - 1;
		int Q = s1.nextInt();

		int colors[] = s1.nextIntArrayOneBased(V);
		adj = new ArrayList[V + 1];
		for(int i=1;i<=V;i++)
			adj[i] = new ArrayList<>();

		while(E-->0) {
			int u = s1.nextInt();
			int v = s1.nextInt();
			adj[u].add(v);
			adj[v].add(u);
		}

		L = new int[V + 1];
		R = new int[V + 1];
		newToOld = new int[V + 1];

		dfs(1, 0);

		SegmentTreeLazyProp segmentTree = new SegmentTreeLazyProp(colors);

		while(Q-->0) {
			int type = s1.nextInt();
			if(1 == type)
				segmentTree.update(s1.nextInt(), s1.nextInt());
			else
				out.println(segmentTree.query(s1.nextInt()));
		}

	}



	/************************ SOLUTION ENDS HERE ************************/





	/************************ TEMPLATE STARTS HERE *********************/

	public static void main(String []args) throws IOException {
		FastScanner in  = new FastScanner(System.in);
		PrintWriter out = 
				new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false); 
		solve(in, out);
		in.close();
		out.close();
	}    

	static class FastScanner{
		BufferedReader reader;
		StringTokenizer st;
		FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}	
		String next()
		{while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}		    
		st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
		String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}	    	  	
		int    nextInt()   {return Integer.parseInt(next());}
		long   nextLong()  {return Long.parseLong(next());}		
		double nextDouble(){return Double.parseDouble(next());}
		char   nextChar()  {return next().charAt(0);}
		int[]  nextIntArray(int n)         {int[] a= new int[n];   int i=0;while(i<n){a[i++]=nextInt();}  return a;}
		long[] nextLongArray(int n)        {long[]a= new long[n];  int i=0;while(i<n){a[i++]=nextLong();} return a;}	
		int[]  nextIntArrayOneBased(int n) {int[] a= new int[n+1]; int i=1;while(i<=n){a[i++]=nextInt();} return a;}	    	
		long[] nextLongArrayOneBased(int n){long[]a= new long[n+1];int i=1;while(i<=n){a[i++]=nextLong();}return a;}	    	
		void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}				
	}

	/************************ TEMPLATE ENDS HERE ************************/
}