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:
// created on : Wed Jul 22 2020
 
import java.io.*;
import java.util.*;

public class E442D2 {

    public static void main(String[] args) throws IOException {
        Solution s = new Solution();
        s.solve();
        s.close();
    }
}

class Solution {
    int[] euler, start, end;
    int time = 0;
    
    void dfs(List<Integer>[] adj, int u, int p){
        start[u] = time;
        euler[time++] = u;
        for(int v: adj[u]){
            if(v == p){
                continue;
            }
            dfs(adj, v, u);
        }
        end[u] = time;
        euler[time++] = u;
    }

	void solve() throws IOException{
        int n = in.nextInt();
        List<Integer>[] adj = new ArrayList[n];
        for(int i = 0; i < n; i++){
            adj[i] = new ArrayList<>();
        }
        for(int i = 0; i < n - 1; i++){
            int p = in.nextInt() - 1;
            adj[i + 1].add(p);
            adj[p].add(i + 1);
        }
        
        int N = 2 * n;
        euler = new int[N];
        start = new int[n];
        end = new int[n];
        dfs(adj, 0, - 1);

        int[] t = readInt(n);
        int[] light = new int[N];
        for(int i = 0; i < N; i++){
            light[i] = t[euler[i]];
        }

        LazySegmentTree tree = new LazySegmentTree(light, N);

        int q = in.nextInt();
        while(q-->0){
            String task = in.next();
            int v = in.nextInt() - 1;
            if(task.equals("get")){
                println(tree.query(0, 0, N - 1, start[v], end[v]) / 2);
            }
            else{
                tree.updateRange(0, 0, N - 1, start[v], end[v]);
            }
        }
	}

	static final int MIN = Integer.MIN_VALUE;
    static final int MAX = Integer.MAX_VALUE;
    static final Reader in = new Reader();
    static final PrintWriter out = new PrintWriter(System.out);


    int[] readInt(int N) throws IOException { 
    	int[] arr = new int[N];
    	for(int i = 0; i < N; i++){ 
    		arr[i] = in.nextInt(); 
    	} 
    	return arr;
    }
    
    long[] readLong(int N) throws IOException { 
    	long[] arr = new long[N];
    	for(int i = 0; i < N; i++){ 
    		arr[i] = in.nextLong(); 
    	} 
    	return arr;
    }
    
    void print(Object O) {  
    	out.print(O); 
    }
    
    void println(Object O) { 
    	out.println(O); 
    }
    
    void println(int[] arr) { 
    	for(int i = 0; i < arr.length; i++){ 
    		print(arr[i] + " "); 
    	} 
    	println(""); 
    }
    
    void displayRuntime(long startTime){
        long stopTime = System.currentTimeMillis();
        long elapsedTime = stopTime - startTime;
        System.out.printf("Runtime: %.2f /n", elapsedTime / 1000D);
    }

    void close() throws IOException{
        in.close();
        out.close();
    }
}

class LazySegmentTree { 
	int size;
	int tree[];
	int lazy[];

	LazySegmentTree(int a[], int n){
		// size of segment tree = 2 * 2 ^ ceil(log(n)) - 1
		size = (int) (2 * Math.pow(2.0, Math.floor((Math.log((double) n) / Math.log(2.0)) + 1)));
		tree = new int[size];
		lazy = new int[size];
		build(a, 0, 0, n - 1);
	}

	// linear tree construction
	void build(int a[], int idx, int low, int high){
		// leaf node
		if(low == high){
			tree[idx] = a[low]; 
		}
		else{
			int mid = low + (high - low) / 2;
			build(a, 2*idx + 1, low, mid);
			build(a, 2*idx + 2, mid + 1, high);
			tree[idx] = tree[2*idx + 1] + tree[2*idx + 2];
		}
	}

	// we just want to toggle all the bulbs
	void updateRange(int idx, int low, int high, int i, int j) { 
		if (lazy[idx] != 0) { 
            // invert the number of on bulbs
			tree[idx] = (high - low + 1) - tree[idx]; 
			if (low != high) { 
				lazy[2*idx + 1] ^= 1; 
				lazy[2*idx + 2] ^= 1; 
			} 
			lazy[idx] = 0; 
		} 

		// out of range 
		if (low > high || low > j || high < i) {
			return; 
		}

		// complete overlap
		if (low >= i && high <= j) {  
            // invert the number of on bulbs
			tree[idx] = (high - low + 1) - tree[idx]; 
			if (low != high) {
				lazy[2 * idx + 1] ^= 1; 
				lazy[2 * idx + 2] ^= 1; 
			} 
		} 
		else{
			// partial overlap
			int mid = low + (high - low) / 2; 
			updateRange(2*idx + 1, low, mid, i, j); 
			updateRange(2*idx + 2, mid + 1, high, i, j); 
			tree[idx] = tree[2*idx + 1] + tree[2*idx + 2]; 
		}
	} 

	// query(0, 0, n - 1, i, j)
	int query(int idx, int low, int high, int i, int j){ 
		if (lazy[idx] != 0) { 
            // invert the number of on bulbs
			tree[idx] = (high - low + 1) - tree[idx]; 
			if (low != high) { 
				lazy[2*idx + 1] ^= 1; 
				lazy[2*idx + 2] ^= 1; 
			} 
			lazy[idx] = 0; 
		} 

		// out of range 
		if (low > high || low > j || high < i) {
			return 0; 
		}
		// complete overlap
		if (low >= i && high <= j) {
			return tree[idx]; 
		}
		// partial overlap
		int mid = low + (high - low) / 2; 
		return query(2*idx + 1, low, mid, i, j) + query(2*idx + 2, mid + 1, high, i, j); 
	} 
}

class Reader {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String next() throws IOException {
        while (!st.hasMoreTokens() ) { 
            st = new StringTokenizer(br.readLine()); 
        }
        return st.nextToken();
    }

    int nextInt() throws IOException { 
        return Integer.parseInt(next()); 
    }
    
    long nextLong() throws IOException { 
        return Long.parseLong(next()); 
    }

    double nextDouble() throws IOException { 
        return Double.parseDouble(next());
    }
    
    String nextLine() throws IOException { 
        return br.readLine(); 
    }
    
    void close() throws IOException { 
        br.close(); 
    }
}