Problem: You are given a tree of n nodes. The tree is rooted at node 1, which is not considered as a leaf regardless of its degree.

Each leaf of the tree has one of the two colors: red or blue. Leaf node v initially has color s_{v}.

The color of each of the internal nodes (including the root) is determined as follows. 

  * Let b be the number of blue immediate children, and r be the number of red immediate children of a given vertex. 
  * Then the color of this vertex is blue if and only if b - r ≥ k, otherwise red. 



Integer k is a parameter that is same for all the nodes.

You need to handle the following types of queries: 

  * 1 v: print the color of node v; 
  * 2 v c: change the color of leaf v to c (c = 0 means red, c = 1 means blue); 
  * 3 h: update the current value of k to h. 

Input

The first line of the input consists of two integers n and k (2 ≤ n ≤ 10^{5}, -n ≤ k ≤ n) — the number of nodes and the initial parameter k.

Each of the next n - 1 lines contains two integers u and v (1 ≤ u,v ≤ n), denoting that there is an edge between vertices u and v.

The next line consists of n space separated integers — the initial array s (-1 ≤ s_i ≤ 1). s_{i} = 0 means that the color of node i is red. s_{i} = 1 means that the color of node i is blue. s_{i} = -1 means that the node i is not a leaf.

The next line contains an integer q (1 ≤ q ≤ 10^5), the number of queries.

q lines follow, each containing a query in one of the following queries: 

  * 1 v (1 ≤ v ≤ n): print the color of node v; 
  * 2 v c (1 ≤ v ≤ n, c = 0 or c = 1): change the color of leaf v to c (c = 0 means red, c = 1 means blue). It is guaranteed that v is a leaf; 
  * 3 h (-n ≤ h ≤ n): update the current value of k to h. 

Output

For each query of the first type, print 0 if the color of vertex v is red, and 1 otherwise.

Example

Input


5 2
1 2
1 3
2 4
2 5
-1 -1 0 1 0
9
1 1
1 2
3 -2
1 1
1 2
3 1
2 5 1
1 1
1 2


Output


0
0
1
1
0
1

Note

Figures:

(i) The initial tree 

(ii) The tree after the 3rd query 

(iii) The tree after the 7th query 

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 400005;
using pii = pair<int, int>;
using ll = long long;
int n, kk, q, sz[N], fa[N], ans[N], clr[N];
vector<int> v[N];
void dfs(int pos, int f) {
  sz[pos] = 1;
  fa[pos] = f;
  for (auto &i : v[pos])
    if (i != f) dfs(i, pos), sz[pos] += sz[i];
}
int op[N][3];
int sum[N];
bool lim[N];
vector<pii> qq[N];
void calc() {
  priority_queue<pii> pq;
  for (int i = 1; i <= n; i++)
    if (clr[i] == -1) clr[i] = 1;
  for (int i = 1; i <= n; i++)
    for (auto &j : v[i])
      if (j != fa[i]) sum[i] += 2 * clr[j] - 1;
  for (int i = 1; i <= n; i++)
    if (v[i].size() - !!fa[i]) pq.push({-sum[i] - 1, i});
  for (int i = -n; i <= n; i++) {
    while (!pq.empty() && -pq.top().first <= i) {
      int nd = pq.top().second;
      pq.pop();
      clr[nd] = 0;
      lim[nd] = 1;
      if (fa[nd] && !lim[fa[nd]]) pq.push({-(sum[fa[nd]] -= 2) - 1, fa[nd]});
    }
    for (auto &j : qq[n + i]) ans[j.second] = 2 * clr[j.first] - 1;
  }
}
int calc(int pos) {
  if (clr[pos] != -1) return clr[pos];
  int sum = 0;
  for (int i = 0;
       i < int(v[pos].size()) && !(sum + int(v[pos].size()) - i < kk ||
                                   sum - int(v[pos].size()) + i >= kk);
       i++)
    if (v[pos][i] != fa[pos]) sum += 2 * calc(v[pos][i]) - 1;
  return sum >= kk;
}
void solve() {
  for (int i = 1; i <= n; i++)
    sort(v[i].begin(), v[i].end(), [](int a, int b) { return sz[a] < sz[b]; });
  for (int i = 1; i <= q; i++) {
    int tp = op[i][0], x = op[i][1];
    if (tp == 1)
      cout << calc(x) << endl;
    else if (tp == 2)
      clr[x] = op[i][2];
    else
      kk = x;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> kk;
  int t1, t2;
  for (int i = 1; i < n; i++) {
    cin >> t1 >> t2;
    v[t1].push_back(t2);
    v[t2].push_back(t1);
  }
  for (int i = 1; i <= n; i++) cin >> clr[i];
  dfs(1, 0);
  bool fl = 1;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> op[i][0] >> op[i][1];
    if (op[i][0] == 2) {
      cin >> op[i][2];
      fl = 0;
    }
  }
  if (fl) {
    for (int i = 1; i <= q; i++) {
      int tp = op[i][0], x = op[i][1];
      if (tp == 3)
        kk = x;
      else
        qq[kk + n].push_back({x, i});
    }
    calc();
    for (int i = 1; i <= q; i++)
      if (ans[i]) cout << (ans[i] + 1) / 2 << endl;
    return 0;
  }
  solve();
  return 0;
}