Polycarpus got hold of a family tree. The found tree describes the family relations of n people, numbered from 1 to n. Every person in this tree has at most one direct ancestor. Also, each person in the tree has a name, the names are not necessarily unique.

We call the man with a number a a 1-ancestor of the man with a number b, if the man with a number a is a direct ancestor of the man with a number b.

We call the man with a number a a k-ancestor (k > 1) of the man with a number b, if the man with a number b has a 1-ancestor, and the man with a number a is a (k - 1)-ancestor of the 1-ancestor of the man with a number b.

In the tree the family ties do not form cycles. In other words there isn't a person who is his own direct or indirect ancestor (that is, who is an x-ancestor of himself, for some x, x > 0).

We call a man with a number a the k-son of the man with a number b, if the man with a number b is a k-ancestor of the man with a number a.

Polycarpus is very much interested in how many sons and which sons each person has. He took a piece of paper and wrote m pairs of numbers vi, ki. Help him to learn for each pair vi, ki the number of distinct names among all names of the ki-sons of the man with number vi.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. Next n lines contain the description of people in the tree. The i-th line contains space-separated string si and integer ri (0 ≤ ri ≤ n), where si is the name of the man with a number i, and ri is either the number of the direct ancestor of the man with a number i or 0, if the man with a number i has no direct ancestor. 

The next line contains a single integer m (1 ≤ m ≤ 105) — the number of Polycarpus's records. Next m lines contain space-separated pairs of integers. The i-th line contains integers vi, ki (1 ≤ vi, ki ≤ n).

It is guaranteed that the family relationships do not form cycles. The names of all people are non-empty strings, consisting of no more than 20 lowercase English letters.

Output

Print m whitespace-separated integers — the answers to Polycarpus's records. Print the answers to the records in the order, in which the records occur in the input.

Examples

Input

6
pasha 0
gerald 1
gerald 1
valera 2
igor 3
olesya 1
5
1 1
1 2
1 3
3 1
6 1


Output

2
2
0
1
0


Input

6
valera 0
valera 1
valera 1
gerald 0
valera 4
kolya 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

1
0
0
0
2
0
0
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.io.BufferedReader;


public class fstclass {

	static boolean[] used = new boolean[300000];
	static ArrayList<Integer>[] list = new ArrayList[300000];
	static ArrayList<Integer>[] d = new ArrayList[300000];
	static int tin[] = new int[300000];
	static int tout[] = new int[300000];
	static int id[] = new int[300000];
	static int timer = 2;
	public static void main(String[] args) {
		InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        int n = in.nextInt();
        int x;
        for(int i = 0; i < 300000; i ++) {
        	list[i] = new ArrayList<>();
        	d[i] = new ArrayList<>();
        	d[i].add(n + 1);
        }
        tin[n + 1] = 0;
        tout[n + 1] = 0;
        for(int i = 1; i <= n; i ++) {
        	x = in.nextInt();
        	list[x].add(i);
        }
        dfs(0, 0);
        int m = in.nextInt();
        int v, p;
        for(int i = 1; i <= m; i ++) {
        	v = in.nextInt();
        	p = in.nextInt();
        	if(id[v] > p) {
        		int ind = id[v] - p;
        		int l = 0, r = d[ind].size();
        		while(r - l > 1) {
        			int mid = (l + r) / 2;
        			if(tin[d[ind].get(mid)] > tin[v])
        				r = mid;
        			else
        				l = mid;
        		}
        		int l1 = 0, r1 = d[id[v]].size();
        		while(r1 - l1 > 1) {
        			int mid = (l1 + r1) / 2;
        			if(tin[d[id[v]].get(mid)] > tin[d[ind].get(l)])
        				r1 = mid;
        			else
        				l1 = mid;
        		}
        		int l2 = 0, r2 = d[id[v]].size();
        		while(r2 - l2 > 1) {
        			int mid = (l2 + r2) / 2;
        			if(tout[d[id[v]].get(mid)] > tout[d[ind].get(l)])
        				r2 = mid;
        			else
        				l2 = mid;
        		}
        		System.out.print(l2 - r1 + " ");
        	}
        	else {
        		System.out.print("0 ");
        	}
        }
	}
	public static void dfs(int x, int s) {
		used[x] = true;
		tin[x] = ++timer;
		for(int i = 0; i < list[x].size(); i ++) {
			int to = list[x].get(i);
			if(!used[to]) {
				dfs(to, s + 1);
			}
		}
		tout[x] = ++timer;
		d[s].add(x);
		id[x] = s;
	}
	
	
	
	
	
	
	
	
	static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            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());
        }

    }

}
