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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author ATailouloute
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        QuickScanner in = new QuickScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        int n;
        int m;
        int u;
        int v;
        int time = 0;
        int[] first_index;
        int[] last_index;
        int[] c;
        int[] tt;
        List<Integer>[] tree;
        boolean[] visited;

        public void solve(int testNumber, QuickScanner in, PrintWriter out) {
            n = in.nextInt();
            m = in.nextInt();
            c = in.nextIntArray(n);
            tree = new List[n];
            first_index = new int[n];
            last_index = new int[n];
            visited = new boolean[n];
            tt = new int[8 * n];
            for (int i = 0; i < n; i++) {
                tree[i] = new ArrayList<>();
            }
            for (int i = 1; i < n; i++) {
                u = in.nextInt() - 1;
                v = in.nextInt() - 1;
                tree[u].add(v);
                tree[v].add(u);
            }
            dfs_order(0);
            init();

            SegmentTree<Long> st = new SegmentTree<Long>(time) {
                protected Long getUpdatedValue(Long oldValue, Long value) {
                    return value;
                }

                protected Long join(Long left, Long right) {
                    return left | right;
                }

                protected Long initNode(int index) {
                    return 1L << c[tt[index]];
                }

                protected Long neutralValue() {
                    return 0L;
                }
            };

            for (int i = 0; i < m; i++) {
                int t = in.nextInt();
                int v = in.nextInt() - 1;
                if (t == 1) {
                    int color = in.nextInt();
                    st.update(first_index[v], last_index[v], 1L << color);
                } else {
                    out.println(Long.bitCount(st.query(first_index[v], last_index[v])));
                }
            }
        }

        void dfs_order(int node) {
            tt[time++] = node;
            visited[node] = true;
            for (int child : tree[node]) {
                if (!visited[child]) {
                    dfs_order(child);
                }
            }
            tt[time++] = node;
        }

        void init() {
            Arrays.fill(first_index, -1);
            for (int i = 0; i < time; i++) {
                if (first_index[tt[i]] == -1) {
                    first_index[tt[i]] = i;
                }
                last_index[tt[i]] = i;
            }
        }

    }

    static class QuickScanner {
        BufferedReader br;
        StringTokenizer st;
        InputStream is;

        public QuickScanner(InputStream stream) {
            is = stream;
            br = new BufferedReader(new InputStreamReader(stream), 32768);
        }

        String nextToken() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(nextToken());
        }

        int[] nextIntArray(int len) {
            int[] a = new int[len];
            for (int i = 0; i < len; i++) {
                a[i] = nextInt();
            }
            return a;
        }

    }

    static abstract class SegmentTree<T> {
        protected int size;
        protected T[] st;
        protected T[] lazy;

        public SegmentTree(int size) {
            this.size = size;
            int maxNodes = Math.max(1, Integer.highestOneBit(size) << 2);
            st = (T[]) new Object[maxNodes];
            lazy = (T[]) new Object[maxNodes];

            build(1, 0, size - 1);
        }

        private void build(int node, int left, int right) {
            if (left == right) {
                st[node] = initNode(left);
            } else {
                int middle = (left + right) >> 1;
                build(node << 1, left, middle);
                build(node << 1 | 1, middle + 1, right);
                st[node] = join(st[node << 1], st[node << 1 | 1]);
            }
        }

        public T query(int from, int to) {
            return query(1, 0, size - 1, from, to);
        }

        private T query(int node, int left, int right, int from, int to) {
            if (left > right || left > to || right < from)
                return neutralValue();

            push(node, left, right);

            if (left >= from && right <= to) {
                return st[node];
            }

            int middle = (left + right) >> 1;
            T leftResult = query(node << 1, left, middle, from, to);
            T rightResult = query(node << 1 | 1, middle + 1, right, from, to);

            return join(leftResult, rightResult);
        }

        public void update(int from, int to, T value) {
            update(1, 0, size - 1, from, to, value);
        }

        protected void update(int node, int left, int right, int from, int to, T value) {
            push(node, left, right);
            if (left > right || left > to || right < from) return;
            if (left >= from && right <= to) {
                st[node] = value;
                if (left != right) {
                    lazy[node << 1] = lazy[node << 1 | 1] = value;
                }
                lazy[node] = null;
                return;
            }
            int middle = (left + right) >> 1;
            update(node << 1, left, middle, from, to, value);
            update(node << 1 | 1, middle + 1, right, from, to, value);
            st[node] = join(st[node << 1], st[node << 1 | 1]);
        }

        protected void push(int node, int left, int right) {
            if (lazy[node] != null) {
                st[node] = getUpdatedValue(st[node], lazy[node]);
                if (left != right) {
                    lazy[node << 1] = lazy[node << 1 | 1] = lazy[node];
                }
                lazy[node] = null;
            }
        }

        protected abstract T getUpdatedValue(T oldValue, T value);

        protected abstract T join(T left, T right);

        protected abstract T initNode(int index);

        protected abstract T neutralValue();

    }
}