Problem: Danil decided to earn some money, so he had found a part-time job. The interview have went well, so now he is a light switcher.

Danil works in a rooted tree (undirected connected acyclic graph) with n vertices, vertex 1 is the root of the tree. There is a room in each vertex, light can be switched on or off in each room. Danil's duties include switching light in all rooms of the subtree of the vertex. It means that if light is switched on in some room of the subtree, he should switch it off. Otherwise, he should switch it on.

Unfortunately (or fortunately), Danil is very lazy. He knows that his boss is not going to personally check the work. Instead, he will send Danil tasks using Workforces personal messages.

There are two types of tasks: 

  1. pow v describes a task to switch lights in the subtree of vertex v.
  2. get v describes a task to count the number of rooms in the subtree of v, in which the light is turned on. Danil should send the answer to his boss using Workforces messages.



A subtree of vertex v is a set of vertices for which the shortest path from them to the root passes through v. In particular, the vertex v is in the subtree of v.

Danil is not going to perform his duties. He asks you to write a program, which answers the boss instead of him.

Input

The first line contains a single integer n (1 ≤ n ≤ 200 000) — the number of vertices in the tree.

The second line contains n - 1 space-separated integers p2, p3, ..., pn (1 ≤ pi < i), where pi is the ancestor of vertex i.

The third line contains n space-separated integers t1, t2, ..., tn (0 ≤ ti ≤ 1), where ti is 1, if the light is turned on in vertex i and 0 otherwise.

The fourth line contains a single integer q (1 ≤ q ≤ 200 000) — the number of tasks.

The next q lines are get v or pow v (1 ≤ v ≤ n) — the tasks described above.

Output

For each task get v print the number of rooms in the subtree of v, in which the light is turned on.

Example

Input

4
1 1 1
1 0 0 1
9
get 1
get 2
get 3
get 4
pow 1
get 1
get 2
get 3
get 4


Output

2
0
0
1
2
1
1
0

Note

<image> The tree before the task pow 1.

<image> The tree after the task pow 1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
template <typename A>
ostream& operator<<(ostream& cout, vector<A> const& v);
template <typename A, typename B>
ostream& operator<<(ostream& cout, pair<A, B> const& p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream& operator<<(ostream& cout, vector<A> const& v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream& operator>>(istream& cin, pair<A, B>& p) {
  cin >> p.first;
  return cin >> p.second;
}
mt19937 rng(steady_clock::now().time_since_epoch().count());
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long double pi = 3.14159265358979323846;
const long long mod = 1000000007;
template <typename num_t>
struct segtree {
  int n, depth;
  vector<num_t> tree, lazy;
  void init(int second, long long* arr) {
    n = second;
    tree = vector<num_t>(4 * second, 0);
    lazy = vector<num_t>(4 * second, 0);
    init(0, 0, n - 1, arr);
  }
  num_t init(int i, int l, int r, long long* arr) {
    if (l == r) return tree[i] = arr[l];
    int mid = (l + r) / 2;
    num_t a = init(2 * i + 1, l, mid, arr),
          b = init(2 * i + 2, mid + 1, r, arr);
    return tree[i] = a.op(b);
  }
  void update(int l, int r, num_t v) {
    if (l > r) return;
    update(0, 0, n - 1, l, r, v);
  }
  num_t update(int i, int tl, int tr, int ql, int qr, num_t v) {
    eval_lazy(i, tl, tr);
    if (tl > tr || tr < ql || qr < tl) return tree[i];
    if (ql <= tl && tr <= qr) {
      lazy[i] = lazy[i].val + v.val;
      eval_lazy(i, tl, tr);
      return tree[i];
    }
    if (tl == tr) return tree[i];
    int mid = (tl + tr) / 2;
    num_t a = update(2 * i + 1, tl, mid, ql, qr, v),
          b = update(2 * i + 2, mid + 1, tr, ql, qr, v);
    return tree[i] = a.op(b);
  }
  num_t query(int l, int r) {
    if (l > r) return num_t::null_v;
    return query(0, 0, n - 1, l, r);
  }
  num_t query(int i, int tl, int tr, int ql, int qr) {
    eval_lazy(i, tl, tr);
    if (ql <= tl && tr <= qr) return tree[i];
    if (tl > tr || tr < ql || qr < tl) return num_t::null_v;
    int mid = (tl + tr) / 2;
    num_t a = query(2 * i + 1, tl, mid, ql, qr),
          b = query(2 * i + 2, mid + 1, tr, ql, qr);
    return a.op(b);
  }
  void eval_lazy(int i, int l, int r) {
    tree[i] = (lazy[i].val & 1) ? ((r - l + 1) - tree[i].val) : tree[i];
    if (l != r) {
      lazy[i * 2 + 1] = (lazy[i].val + lazy[i * 2 + 1].val) & 1;
      lazy[i * 2 + 2] = (lazy[i].val + lazy[i * 2 + 2].val) & 1;
    }
    lazy[i] = num_t();
  }
};
struct sum_t {
  long long val;
  static const long long null_v = 0;
  sum_t() : val(0) {}
  sum_t(long long v) : val(v) {}
  sum_t op(sum_t& other) { return sum_t(val + other.val); }
};
long long n, m, k, q, Q, T, l, r, x, y, z;
long long a[1000005];
long long b[1000005];
long long c[1000005];
string second, t;
long long ans = 0;
vector<int> edges[200005];
int st[200005], en[200005];
int TIME = -1;
segtree<sum_t> sst;
void dfs_t(int v) {
  st[v] = ++TIME;
  for (int x : edges[v]) dfs_t(x);
  en[v] = TIME;
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  };
  cin >> n;
  for (long long i = 0; i < (n - 1); i++) {
    cin >> x;
    edges[--x].push_back(i + 1);
  }
  dfs_t(0);
  for (int ele = 0; ele < n; ele++) cin >> a[ele];
  ;
  memset(b, 0, sizeof(b));
  sst.init(n, b);
  for (long long i = 0; i < (n); i++) sst.update(st[i], st[i], a[i]);
  cin >> q;
  for (long long i = 0; i < (q); i++) {
    cin >> second >> x;
    --x;
    if (second == "pow")
      sst.update(st[x], en[x], 1);
    else
      cout << sst.query(st[x], en[x]).val << '\n';
  }
}