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.util.*;
import java.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=1000000007;
	long gcd(long a,long b) {
		return a==0?b:b>=a?gcd(b%a,a):gcd(b,a);
	}
	ArrayList<Integer>[] graph;
	int[][] A;
	void work() {
		int n=in.nextInt();
		graph=(ArrayList<Integer>[])new ArrayList[n];
		A=new int[n][2];
		for(int i=0;i<n;i++)graph[i]=new ArrayList<>();
		for(int i=1;i<n;i++) {
			int s=in.nextInt()-1;
			graph[s].add(i);
			graph[i].add(s);
		}
		dfs(0,new int[1],new boolean[n]);
		Node root=new Node();
		for(int i=0;i<n;i++) {
			int v=in.nextInt();
			if(v==1) {
				insert(root,0,n,A[i][0]);
			}
		}
		int q=in.nextInt();
		for(int i=0;i<q;i++) {
			String s=in.next();
			int t=1;
			if(s.equals("pow")) {
				t=2;
			}
			int node=in.nextInt()-1;
			if(t==1) {
				out.println(search(root,0,n,A[node][0],A[node][1]));
			}else {
				update(root,0,n,A[node][0],A[node][1]);
			}
		}
	}
	private void update(Node node, int l, int r, int s, int e) {
		updatelazy(node,l,r);
		if(s<=l&&r<=e) {
			node.sum=r-l+1-node.sum;
			if(l!=r) {
				getLnode(node).lazy^=1;
				getRnode(node).lazy^=1;
			}
			return;
		}
		int m=l+(r-l)/2;
		if(m>=s) {
			update(getLnode(node),l,m,s,e);
		}
		if(m+1<=e) {
			update(getRnode(node),m+1,r,s,e);
		}
		updatelazy(getLnode(node),l,m);
		updatelazy(getRnode(node),m+1,r);
		node.sum=getLnode(node).sum+getRnode(node).sum;//左右节点可能有lazy标记
	}
	void updatelazy(Node node,int l,int r) {
		if(node.lazy==1) {
			node.sum=(r-l+1)-node.sum;
			node.lazy=0;
			if(l!=r) {
				getLnode(node).lazy^=1;
				getRnode(node).lazy^=1;
			}
		}
	}
	private int search(Node node, int l, int r, int s, int e) {
		updatelazy(node,l,r);
		if(s<=l&&r<=e) {
			return node.sum;
		}
		int m=l+(r-l)/2;
		int ret=0;
		if(m>=s) {
			ret+=search(getLnode(node),l,m,s,e);
		}
		if(m+1<=e) {
			ret+=search(getRnode(node),m+1,r,s,e);
		}
		return ret;
	}
	private void insert(Node node,int l,int r,int idx) {
		if(l==r) {
			node.sum=1;
			return;
		}
		Node lnode=getLnode(node);
		Node rnode=getRnode(node);
		int m=l+(r-l)/2;
		if(idx<=m) {
			insert(lnode,l,m,idx);
		}else {
			insert(rnode,m+1,r,idx);
		}
		node.sum=lnode.sum+rnode.sum;
	}
	private Node getLnode(Node node) {
		if(node.lnode==null)node.lnode=new Node();
		return node.lnode;
	}
	private Node getRnode(Node node) {
		if(node.rnode==null)node.rnode=new Node();
		return node.rnode;
	}
	private void dfs(int node, int[] B,boolean[] vis) {
		A[node][0]=B[0];
		vis[node]=true;
		for(int nn:graph[node]) {
			if(!vis[nn]) {
				B[0]++;
				dfs(nn,B,vis);
			}
		}
		A[node][1]=B[0];
	}
	class Node{
		Node lnode,rnode;
		int lazy,sum;
	}
}	



class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}


	public String next() 
	{
		while(st==null || !st.hasMoreElements())//回车，空行情况
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	public int nextInt() 
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}