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.text.*;
import java.util.*;
import java.math.*;
public class template {
	public static void main(String[] args) throws Exception {
		new template().run();
	}
	int[] arr;
	int cur;
	int[] lazy;
	int[] vals;
	public void run() throws Exception {
		FastScanner f = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int ncnt = f.nextInt();
		int[] st = new int[ncnt], end = new int[ncnt];
		TreeNode[] tn = new TreeNode[ncnt];
		arr = new int[ncnt];
		for(int i = 0; i < ncnt; i++) 
			tn[i] = new TreeNode(i);
		for(int i = 1; i < ncnt; i++)
			tn[f.nextInt()-1].children.add(tn[i]);
		for(int i = 0; i < ncnt; i++)
			tn[i].val = f.nextInt();
		tn[0].traverse(st, end);
		lazy = new int[4*ncnt];
		vals = new int[4*ncnt];
		//System.out.println(Arrays.toString(st));
		//System.out.println(Arrays.toString(end));
		build(1,0,ncnt);
		int qcnt = f.nextInt();
		while(qcnt-->0) {
			String s = f.next();
			int i =f.nextInt()-1;
			if(s.equals("get")) {
				out.println(query(1, 0, ncnt, st[i], end[i]));
			} else {
				update(1,0,ncnt,st[i], end[i]);
			}
			//print(1, 0, ncnt);
			//System.out.println();
		}
		out.flush();
	}
	public void print(int i, int l, int r) {
		System.out.println(i + " " + l + "-" + r + " val:" + vals[i] + " lazy:" + lazy[i]);
		if(l != r-1) {
			int m = (l+r)/2;
			print(2*i, l, m);
			print(2*i+1, m, r);
		}
	}
	public int query(int i, int l, int r, int tl, int tr) {
		if(r <= tl || l >= tr) return 0;
		if(lazy[i] != 0) {
			prop(i, l, r);
		}
		if(tr >= r && tl <= l) return vals[i];
		if(l == r-1) return 0;
		int m = (l+r)/2;
		return query(2*i, l, m, tl, tr) + query(2*i+1, m, r, tl, tr);
	}
	public void update(int i, int l, int r, int tl, int tr) {
		if(lazy[i] == 1) prop(i, l, r);
		if(r <= tl || l >= tr) {
			return;
		}
		if(tr >= r && tl <= l) {
			prop(i, l, r);
		}
		else if(l != r-1) {
			int m = (l+r)/2;
			update(2*i, l, m, tl, tr);
			update(2*i+1, m, r, tl, tr);
			vals[i] = vals[2*i] + vals[2*i+1];
		}
	}
	public void prop(int i, int l, int r) {
		lazy[i] = 0;
		vals[i] = r-l-vals[i];
		if(l != r-1) {
			lazy[2*i] ^= 1;
			lazy[2*i+1] ^= 1;
		}
	}
	public void build(int i, int l, int r) {
		if(l == r-1) vals[i] = arr[l];
		else {
			int m = (l+r)/2;
			build(2*i, l, m);
			build(2*i+1, m, r);
			vals[i] = vals[2*i] + vals[2*i+1];
		}
	}
	class TreeNode {
		LinkedList<TreeNode> children;
		int id;
		int val;
		public TreeNode(int n) {
			id = n;
			children = new LinkedList<TreeNode>();
		}
		public void traverse(int[] st, int[] end) {
			st[id] = cur;
			arr[cur] = val;
			cur++;
			for(TreeNode n : children)
				n.traverse(st, end);
			end[id] = cur;
		}
	}
    static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }


    }
}