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.*;
import java.math.BigInteger;
import java.util.*;

public class Main {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(System.out);
    static int MOD = 1000000007;

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        m.solve();
        m.close();
    }

    void close() throws IOException {
        pw.flush();
        pw.close();
        br.close();
    }

    int ri() throws IOException {
        return Integer.parseInt(br.readLine());
    }

    long rl() throws IOException {
        return Long.parseLong(br.readLine());
    }

    int[] ril(int n) throws IOException {
        int[] nums = new int[n];
        int c = 0;
        for (int i = 0; i < n; i++) {
            int sign = 1;
            c = br.read();
            int x = 0;
            if (c == '-') {
                sign = -1;
                c = br.read();
            }
            while (c >= '0' && c <= '9') {
                x = x * 10 + c - '0';
                c = br.read();
            }
            nums[i] = x * sign;
        }
        while (c != '\n' && c != -1) c = br.read();
        return nums;
    }

    long[] rll(int n) throws IOException {
        long[] nums = new long[n];
        int c = 0;
        for (int i = 0; i < n; i++) {
            int sign = 1;
            c = br.read();
            long x = 0;
            if (c == '-') {
                sign = -1;
                c = br.read();
            }
            while (c >= '0' && c <= '9') {
                x = x * 10 + c - '0';
                c = br.read();
            }
            nums[i] = x * sign;
        }
        while (c != '\n' && c != -1) c = br.read();
        return nums;
    }

    char[] rs() throws IOException {
        return br.readLine().toCharArray();
    }

    void solve() throws IOException {
        int[] nm = ril(2);
        int n = nm[0];
        int m = nm[1];
        c = ril(n);  // 1 <= ci <= 60
        adj = new ArrayList<>(n);
        for (int i = 0; i < n; i++) adj.add(new ArrayList<>());
        for (int i = 0; i < n-1; i++) {
            int[] xy = ril(2);
            adj.get(xy[0]-1).add(xy[1]-1);
            adj.get(xy[1]-1).add(xy[0]-1);
        }
        euler = new int[2 * n];
        s = new int[n];
        f = new int[n];
        dfs(0, -1);
        long[] starr = new long[2 * n];
        for (int i = 0; i < euler.length; i++) {
            starr[i] = 1l << c[euler[i]];
        }

        SegmentTree st = new SegmentTree(starr);

        for (int i = 0; i < m; i++) {
            String[] line = br.readLine().split(" ");
            if (Integer.parseInt(line[0]) == 1) {
                int v = Integer.parseInt(line[1]) - 1;
                int c = Integer.parseInt(line[2]);
                int l = s[v];
                int r = f[v];
                st.modify(l, r, 1l << c);
            } else {
                int v = Integer.parseInt(line[1]) - 1;
                pw.println(Long.bitCount(st.query(s[v], f[v])));
            }
        }
    }

    List<List<Integer>> adj;
    int[] c;
    int[] euler;
    int[] s;
    int[] f;
    int idx = 0;
    void dfs(int u, int p) {
        euler[idx] = u;
        s[u] = idx;
        idx++;
        for (int v : adj.get(u)) {
            if (v != p) dfs(v, u);
        }
        euler[idx] = u;
        f[u] = idx;
        idx++;
    }
}

class SegmentTree {
    int n;
    long[] st;
    boolean[] hasLazy;
    long[] lazy;

    // Modify identity and combine together.
    long identity = 0;
    long combine(long a, long b) {
        return a | b;
    }

    // Range assignment overrides the existing value.
    long lazyFunction(long currVal, long lazyVal) {
        return lazyVal;
    }

    SegmentTree(long[] arr) {
        n = arr.length;
        st = new long[4*n];
        hasLazy = new boolean[4*n];
        lazy = new long[4*n];
        build(arr, 1, 0, n-1);
    }

    private void build(long[] arr, int i, int l, int r) {
        if (l == r) st[i] = arr[l];
        else {
            int m = l + (r - l) / 2;
            build(arr, 2*i, l, m);
            build(arr, 2*i+1, m+1, r);
            st[i] = combine(st[2*i], st[2*i+1]);
        }
    }

    private void push(int v) {
        if (!hasLazy[v]) return;
        st[2*v] = lazyFunction(st[2*v], lazy[v]);
        lazy[2*v] = lazyFunction(lazy[2*v], lazy[v]);
        st[2*v+1] = lazyFunction(st[2*v+1], lazy[v]);
        lazy[2*v+1] = lazyFunction(lazy[2*v+1], lazy[v]);
        hasLazy[2*v] = hasLazy[2*v+1] = true;
        hasLazy[v] = false;
    }

    void modify(int l, int r, long value) {
        modify(1, 0, n-1, l, r, value);
    }

    private void modify(int v, int l, int r, int ql, int qr, long value) {
        if (ql > qr) return;
        if (l == ql && r == qr) {
            st[v] = lazyFunction(st[v], value);
            lazy[v] = lazyFunction(lazy[v], value);
            hasLazy[v] = true;
        } else {
            push(v);
            int m = l + (r - l) / 2;
            modify(2*v, l, m, ql, Math.min(qr, m), value);
            modify(2*v+1, m+1, r, Math.max(ql, m+1), qr, value);
            st[v] = combine(st[2*v], st[2*v+1]);
        }
    }

    // Note: input range is closed [l, r]
    long query(int l, int r) {
        return query(1, 0, n-1, l, r);
    }

    private long query(int v, int l, int r, int ql, int qr) {
        // System.out.println(l + " " + r + " " + ql + " " + qr);
        if (ql > qr) return identity;
        if (l == ql && r == qr) return st[v];
        push(v);
        int m = l + (r - l) / 2;
        long resl = query(2*v, l, m, ql, Math.min(qr, m));
        long resr = query(2*v+1, m+1, r, Math.max(ql, m+1), qr);
        return combine(resl, resr);
    }
}