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 warning(disable : 4996)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("-ffloat-store")
using namespace std;
clock_t time_p = clock();
void aryanc403() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || a.first == b.first && a.second <= b.second);
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 1000000007L;
const long long int maxN = 400000L;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m;
string s;
vector<long long int> a, e[maxN + 5], b;
long long int tin[maxN + 5], tout[maxN + 5];
long long int f[4 * maxN + 5], lazy[4 * maxN + 5];
void dfs(long long int u, long long int p) {
  tin[u] = ++T;
  b.push_back(1LL << a[u]);
  for (auto x : e[u]) {
    if (x == p) continue;
    dfs(x, u);
  }
  tout[u] = T;
}
void build(long long int id, long long int l, long long int r) {
  long long int i;
  for (i = (l); i <= (r); ++i) f[id] |= b[i];
  if (l == r) return;
  long long int m = (l + r) / 2;
  build(2 * id, l, m);
  build(2 * id + 1, m + 1, r);
}
void shift(long long int id) {
  if (!lazy[id]) return;
  f[2 * id] = f[2 * id + 1] = lazy[2 * id] = lazy[2 * id + 1] = lazy[id];
  lazy[id] = 0;
}
void update(long long int id, long long int l, long long int r, long long int L,
            long long int R, long long int x) {
  if (r < L || R < l) return;
  if (L <= l && r <= R) {
    f[id] = x;
    lazy[id] = x;
    return;
  }
  long long int m = (l + r) / 2;
  shift(id);
  update(2 * id, l, m, L, R, x);
  update(2 * id + 1, m + 1, r, L, R, x);
  f[id] = f[2 * id] | f[2 * id + 1];
}
long long int query(long long int id, long long int l, long long int r,
                    long long int L, long long int R) {
  if (r < L || R < l) return 0;
  if (L <= l && r <= R) return f[id];
  long long int m = (l + r) / 2;
  shift(id);
  return query(2 * id, l, m, L, R) | query(2 * id + 1, m + 1, r, L, R);
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    cin >> n >> m;
    a.clear();
    a.reserve(n + 1);
    a.push_back(0);
    b.push_back(0);
    for (i = 0; i < (n); ++i) {
      cin >> in;
      a.push_back(in);
    }
    for (i = 0; i < (n - 1); ++i) {
      cin >> u >> v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    dfs(1, 1);
    build(1, 1, n);
    while (m--) {
      cin >> T >> u;
      if (T == 1) {
        cin >> x;
        x = (1LL << x);
        update(1, 1, n, tin[u], tout[u], x);
      } else {
        cout << __builtin_popcountll(query(1, 1, n, tin[u], tout[u])) << "\n";
      }
    }
  }
  aryanc403();
  return 0;
}