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:
/*
Keep solving problems.
*/

import java.util.*;
import java.io.*;

public class CFA {
    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    private static long MOD = 1000L * 1000L * 1000L + 7;
    private static final int[] dx = {0, -1, 0, 1};
    private static final int[] dy = {1, 0, -1, 0};
    private static final String yes = "Yes";
    private static final String no = "No";

    int n;
    int m;
    int[] color;
    List<List<Integer>> graph = new ArrayList<>();
    int[] start;
    int[] end;
    int INDEX = 0;
    long[] tree;
    int[] lazy;
    void build_tree(int node, int nodeStartIndex, int nodeEndIndex) {
        if(nodeStartIndex > nodeEndIndex) return; // Out of range
        if(nodeStartIndex == nodeEndIndex) { // Leaf node
            tree[node] = 1L << color[nodeStartIndex]; // Init value
            return;
        }

        build_tree(node * 2, nodeStartIndex, (nodeStartIndex + nodeEndIndex) / 2); // Init left child
        build_tree(node * 2 + 1, 1 + (nodeStartIndex + nodeEndIndex) / 2, nodeEndIndex); // Init right child
        tree[node] = tree[node * 2] | tree[node * 2 + 1]; // Init root value
    }

    /**
     * Increment elements within range [queryStartIndex, queryEndIndex] with value
     */
    void update_tree(int node, int nodeStartIndex, int nodeEndIndex, int queryStartIndex, int queryEndIndex, int value) {
        if(nodeStartIndex > nodeEndIndex || nodeStartIndex > queryEndIndex || nodeEndIndex < queryStartIndex) // Current segment is not within range [i, j]
            return;
        if(nodeStartIndex >= queryStartIndex && nodeEndIndex <= queryEndIndex) { // Segment is fully within range
            tree[node] = 1L << value;
            lazy[node] = value;
            return;
        }

        //-------------
        shift(node, nodeStartIndex, nodeEndIndex);
        //-------------
        update_tree(node * 2, nodeStartIndex, (nodeStartIndex + nodeEndIndex) / 2, queryStartIndex, queryEndIndex, value); // Updating left child
        update_tree(node * 2 + 1, 1 + (nodeStartIndex + nodeEndIndex) / 2, nodeEndIndex, queryStartIndex, queryEndIndex, value); // Updating right child

        tree[node] = tree[node * 2] | tree[node * 2 + 1]; // Updating root
    }

    /**
     * Query tree to get max element value within range [queryStartIndex, queryEndIndex]
     */
    long query_tree(int node, int nodeStartIndex, int nodeEndIndex, int queryStartIndex, int queryEndIndex) {
        if(nodeStartIndex > nodeEndIndex || nodeStartIndex > queryEndIndex || nodeEndIndex < queryStartIndex) return 0; // Out of range
        if(nodeStartIndex >= queryStartIndex && nodeEndIndex <= queryEndIndex) // Current segment is totally within range [i, j]
            return tree[node];

        //-------------
        shift(node, nodeStartIndex, nodeEndIndex);
        //-------------
        long q1 = query_tree(node * 2, nodeStartIndex, (nodeStartIndex + nodeEndIndex) / 2, queryStartIndex, queryEndIndex); // Query left child
        long q2 = query_tree(1 + node * 2, 1 + (nodeStartIndex + nodeEndIndex) / 2, nodeEndIndex, queryStartIndex, queryEndIndex); // Query right child

        return q1 | q2;
    }

    private void shift(int node, int nodeStartIndex, int nodeEndIndex) {
        if(lazy[node] != 0) { // This node needs to be updated
            if (nodeEndIndex > nodeStartIndex) {
                tree[2 * node] = 1L << lazy[node];
                tree[2 * node + 1] = 1L << lazy[node];

                lazy[node * 2] = lazy[node]; // Mark child as lazy
                lazy[node * 2 + 1] = lazy[node]; // Mark child as lazy
                lazy[node] = 0; // Reset it
            }
        }
    }

    void solve() throws IOException {
        n = nextInt();
        m = nextInt();
        color = nextIntArr(n);
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }

        for (int i = 0; i < n - 1; i++) {
            int u = nextInt() - 1;
            int v = nextInt() - 1;
            graph.get(u).add(v);
            graph.get(v).add(u);
        }

        start = new int[n];
        end = new int[n];
        dfs(-1, 0, start, end);

        tree = new long[5 * n];
        lazy = new int[5 * n];
        for (int i = 0; i < n; i++) {
            update_tree(1, 0, n - 1, start[i], start[i], color[i]);
        }

        for (int i = 0; i < m; i++) {
            int t = nextInt();
            if (t == 1) {
                int v = nextInt() - 1;
                int c = nextInt();
                update_tree(1, 0, n - 1, start[v], end[v], c);
            }
            else {
                int v = nextInt() - 1;
                long val = query_tree(1, 0, n - 1, start[v], end[v]);
                outln(Long.bitCount(val));
            }
        }
    }

    // Euler's tour:
    void dfs(int parent, int nodeIndex, int[] startIndex, int[] endIndex) {
        startIndex[nodeIndex] = INDEX;
        INDEX++;
        for (int nxt : graph.get(nodeIndex)) {
            if (nxt != parent) {
                dfs(nodeIndex, nxt, startIndex, endIndex);
            }
        }

        endIndex[nodeIndex] = INDEX - 1;
    }

    void shuffle(int[] a) {
        int n = a.length;
        for(int i = 0; i < n; i++) {
            int r = i + (int) (Math.random() * (n - i));
            int tmp = a[i];
            a[i] = a[r];
            a[r] = tmp;
        }
    }
    long gcd(long a, long b) {
        while(a != 0 && b != 0) {
            long c = b;
            b = a % b;
            a = c;
        }
        return a + b;
    }
    private void outln(Object o) {
        out.println(o);
    }
    private void out(Object o) {
        out.print(o);
    }
    private void formatPrint(double val) {
        outln(String.format("%.9f%n", val));
    }
    public CFA() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }
    public static void main(String[] args) throws IOException {
        new CFA();
    }

    public long[] nextLongArr(int n) throws IOException{
        long[] res = new long[n];
        for(int i = 0; i < n; i++)
            res[i] = nextLong();
        return res;
    }
    public int[] nextIntArr(int n) throws IOException {
        int[] res = new int[n];
        for(int i = 0; i < n; i++)
            res[i] = nextInt();
        return res;
    }
    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }
    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}