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 JAVA):

Solution:
import java.io.*;
import java.math.*;
import java.util.*;
import java.util.Map.*;

public class Main
{
	static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	static StreamTokenizer sc = new StreamTokenizer(in);
	static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	static Scanner sca = new Scanner(System.in);
	static int inf = Integer.MAX_VALUE;

	static int N = (int) 2e5;

	static ArrayList<Integer> g[] = new ArrayList[N + 10];

	static int arr[] = new int[N + 10];
	static int tree[] = new int[4 * N + 10];
	static int lazy[] = new int[4 * N + 10];

	static int l[] = new int[N + 10];
	static int r[] = new int[N + 10];
	static int idx[] = new int[N + 10];
	static int cnt;

	public static void main(String[] args) throws IOException
	{
// 		System.setIn(new FileInputStream("copycat.in"));// 读入文件
//		System.setOut(new PrintStream(new FileOutputStream("copycat.out")));// 输出到文件

		int n = ini();

		for (int i = 1; i <= n; i++)
		{
			g[i] = new ArrayList<Integer>();
		}
		for (int i = 2; i <= n; i++)
		{
			int p = ini();
			g[p].add(i);
		}
		for (int i = 1; i <= n; i++)
		{
			arr[i] = ini();
		}

		dfs(1);

		build(1, 1, n);

		int q = ini();
		while (q-- > 0)
		{
			String str = ins();
			int v = ini();
			if (str.equals("get"))
			{
				out.println(query(1, 1, n, l[v], r[v]));
			} else if (str.equals("pow"))
			{
				update(1, 1, n, l[v], r[v]);
			}
		}

		out.flush();

	}

	static long query(int node, int start, int end, int left, int right)
	{
		if (left <= start && end <= right)
		{
			return tree[node];
		}
		pushdown(node, start, end);

		int mid = (start + end) >> 1;
		long ans = 0;

		if (left <= mid)
			ans += query(node << 1, start, mid, left, right);
		if (right > mid)
			ans += query(node << 1 | 1, mid + 1, end, left, right);

		return ans;
	}

	static void update(int node, int start, int end, int left, int right)
	{
		if (left <= start && end <= right)
		{
			lazy[node] ^= 1;
			tree[node] = (end - start + 1) - tree[node];
			return;
		}
		pushdown(node, start, end);

		int mid = (start + end) >> 1;
		if (left <= mid)
			update(node << 1, start, mid, left, right);
		if (right > mid)
			update(node << 1 | 1, mid + 1, end, left, right);

		tree[node] = tree[node << 1] + tree[node << 1 | 1];
	}

	static void pushdown(int node, int start, int end)
	{
		if (lazy[node] != 0)
		{
			lazy[node << 1] ^= 1;
			lazy[node << 1 | 1] ^= 1;

			int mid = (start + end) >> 1;

			tree[node << 1] = (mid - start + 1) - tree[node << 1];
			tree[node << 1 | 1] = (end - mid) - tree[node << 1 | 1];

			lazy[node] = 0;
		}
	}

	static void build(int node, int start, int end)
	{
		if (start == end)
		{
			tree[node] = arr[idx[start]];
			return;
		}

		int mid = (start + end) >> 1;

		build(node << 1, start, mid);
		build(node << 1 | 1, mid + 1, end);

		tree[node] = tree[node << 1] + tree[node << 1 | 1];
	}

	static void dfs(int x)
	{
		l[x] = ++cnt;
		idx[cnt] = x;
		for (int i = 0; i < g[x].size(); i++)
		{
			int a = g[x].get(i);
			dfs(a);
		}
		r[x] = cnt;
	}

	static int ini() throws IOException
	{
		sc.nextToken();
		return (int) sc.nval;
	}

	static long inl() throws IOException
	{
		sc.nextToken();
		return (long) sc.nval;
	}

	static double ind() throws IOException
	{
		sc.nextToken();
		return sc.nval;
	}

	static String ins() throws IOException
	{
		sc.nextToken();
		return sc.sval;
	}

	static String rel() throws IOException
	{
		return in.readLine();
	}

}