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.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.StringTokenizer;

public class Solve6 {

    public static void main(String[] args) throws IOException {
        PrintWriter pw = new PrintWriter(System.out);
        new Solve6().solve(pw);
        pw.flush();
        pw.close();
    }

    public void solve(PrintWriter pw) throws IOException {
        FastReader sc = new FastReader();
        int n = sc.nextInt(), m = sc.nextInt();
        c = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            c[i] = sc.nextInt();
        }
        ArrayList<Integer>[] adj = new ArrayList[n + 1];
        for (int i = 1; i <= n; i++) {
            adj[i] = new ArrayList();
        }
        for (int i = 0; i < n - 1; i++) {
            int x = sc.nextInt(), y = sc.nextInt();
            adj[x].add(y);
            adj[y].add(x);
        }
        in = new int[n + 1];
        out = new int[n + 1];
        n <<= 1;
        a = new int[n + 1];
        dfsOrder(1, 0, adj);
        SegmentTree st = new SegmentTree(a, n);
        for (int i = 0; i < m; i++) {
            if (sc.nextInt() == 1) {
                int v = sc.nextInt(), c = sc.nextInt();
                st.update(in[v], out[v], c, 1, n, 1);
            } else {
                int v = sc.nextInt();
                pw.println(st.query(in[v], out[v], 1, n, 1).cardinality());
            }
        }
    }
    int timer;
    int[] in, out, a, c;

    public void dfsOrder(int u, int p, ArrayList<Integer>[] adj) {
        in[u] = ++timer;
        a[in[u]] = c[u];
        for (Integer v : adj[u]) {
            if (v != p) {
                dfsOrder(v, u, adj);
            }
        }
        out[u] = ++timer;
    }

    static class SegmentTree {

        BitSet[] t;
        int[] lazy;

        public SegmentTree(int[] a, int n) {
            t = new BitSet[n << 2];
            lazy = new int[n << 2];
            build(1, n, 1, a);
        }

        public void build(int left, int right, int nd, int[] a) {
            t[nd] = new BitSet();
            if (left == right) {
                if (a[left] != 0) {
                    t[nd].set(a[left]);
                }
                return;
            }
            int mid = (left + right) >> 1;
            build(left, mid, nd << 1, a);
            build(mid + 1, right, nd << 1 | 1, a);
            t[nd].or(t[nd << 1]);
            t[nd].or(t[nd << 1 | 1]);
        }

        public BitSet query(int l, int r, int left, int right, int nd) {
            push(left, right, nd);
            if (l <= left && right <= r) {
                return t[nd];
            }
            if (l > right || r < left) {
                return new BitSet();
            }
            int mid = (left + right) >> 1;
            BitSet x = new BitSet();
            x.or(query(l, r, left, mid, nd << 1));
            x.or(query(l, r, mid + 1, right, nd << 1 | 1));
            return x;
        }

        public void push(int left, int right, int nd) {
            if (lazy[nd] != 0) {
                t[nd].clear();
                t[nd].set(lazy[nd]);
                if (left != right) {
                    lazy[nd << 1] = lazy[nd];
                    lazy[nd << 1 | 1] = lazy[nd];
                }
                lazy[nd] = 0;
            }
        }

        public void update(int l, int r, int c, int left, int right, int nd) {
            push(left, right, nd);
            if (l <= left && right <= r) {
                lazy[nd] = c;
                push(left, right, nd);
                return;
            }
            if (l > right || r < left) {
                return;
            }
            int mid = (left + right) >> 1;
            update(l, r, c, left, mid, nd << 1);
            update(l, r, c, mid + 1, right, nd << 1 | 1);
            t[nd].clear();
            t[nd].or(t[nd << 1]);
            t[nd].or(t[nd << 1 | 1]);
        }
    }

    static class FastReader {

        StringTokenizer st;
        BufferedReader br;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (Exception e) {
            }
        }

        public String next() {
            if (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public boolean hasNext() throws IOException {
            if (st != null && st.hasMoreTokens()) {
                return true;
            }
            String s = br.readLine();
            if (s == null || s.isEmpty()) {
                return false;
            }
            st = new StringTokenizer(s);
            return true;
        }
    }
}