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 CPP):

Solution:
#include <bits/stdc++.h>
#pragma gcc optimize("ofast")
using namespace std;
long long modexp(long long a, long long b, long long m) {
  long long r = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) r = (r * a) % m;
    b = b / 2;
    a = (a * a) % m;
  }
  return r % m;
}
const int N = 1200300;
long long n, m;
long long col[N];
vector<long long> adj[N];
long long Tree[4 * N];
long long convert(long long x) { return (1LL << x); }
long long rangeInd = 0;
long long order[N], leftRange[N], rightRange[N], add[4 * N];
void dfs(long long u, long long p) {
  order[rangeInd] = u;
  leftRange[u] = rangeInd++;
  for (auto v : adj[u]) {
    if (v != p) {
      dfs(v, u);
    }
  }
  rightRange[u] = rangeInd;
}
void build(long long id, long long left, long long right) {
  add[id] = -1;
  if (left + 1 == right) {
    Tree[id] = 1LL << col[order[left]];
  } else {
    long long mid = (left + right) >> 1;
    build(2 * id + 1, left, mid);
    build(2 * id + 2, mid, right);
    Tree[id] = Tree[2 * id + 1] | Tree[2 * id + 2];
  }
}
void push(long long id) {
  if (add[id] == -1) return;
  for (int i = 1; i < 3; i++) {
    Tree[2 * id + i] = add[2 * id + i] = add[id];
  }
  add[id] = -1;
}
void paint(long long id, long long left, long long right, long long leftorder,
           long long rightorder, long long color) {
  if (leftorder >= rightorder) return;
  if (left == leftorder && right == rightorder) {
    Tree[id] = 1LL << color;
    add[id] = 1LL << color;
  } else {
    push(id);
    long long mid = (left + right) >> 1;
    paint(2 * id + 1, left, mid, leftorder, min(mid, rightorder), color);
    paint(2 * id + 2, mid, right, max(mid, leftorder), rightorder, color);
    Tree[id] = Tree[2 * id + 1] | Tree[2 * id + 2];
  }
}
pair<long long, pair<long long, long long>> q[N];
long long get(long long id, long long left, long long right,
              long long leftorder, long long rightorder) {
  if (leftorder >= rightorder) return 0;
  if (left == leftorder && right == rightorder) return Tree[id];
  push(id);
  long long mid = (left + right) >> 1;
  long long ans = 0;
  ans |= get(2 * id + 1, left, mid, leftorder, min(mid, rightorder));
  ans |= get(2 * id + 2, mid, right, max(leftorder, mid), rightorder);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    col[i] = x;
  }
  for (int i = 0; i < n; i++) adj[i].clear();
  for (int i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    --x, --y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 0; i < m; i++) {
    cin >> q[i].first >> q[i].second.first;
    q[i].second.first--;
    if (q[i].first == 1) {
      cin >> q[i].second.second;
    }
  }
  rangeInd = 0;
  dfs(0, -1);
  build(0, 0, n);
  for (int i = 0; i < m; i++) {
    long long types = q[i].first;
    long long node = q[i].second.first;
    if (types == 1) {
      long long c = q[i].second.second;
      paint(0, 0, n, leftRange[node], rightRange[node], c);
    } else {
      long long msk = get(0, 0, n, leftRange[node], rightRange[node]);
      long long ans = 0;
      for (int j = 0; j < 61; j++) {
        ans += ((msk >> j) & 1);
      }
      cout << ans << '\n';
    }
  }
  return 0;
}