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.util.*;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.InputStream;
import java.lang.*;

public class Source {
	public static void main(String[] args) {
		InputReader in = new InputReader(System.in);
		PrintWriter ou = new PrintWriter(System.out);		
		MainCode obj = new MainCode();
		obj.Solve(in, ou);
		ou.close();
	}
	
}

class MainCode {
	private int n, q;
	private int[] h;
	private int[][] p;
	private List<Integer> Root;
	private List<Integer>[] adj;
	
	private int cnt;
	private int[] timeIn, timeOut;
	private List<Integer>[] Vertex;
	
	public void Solve(InputReader inp, PrintWriter out) {
		n 		= inp.nextInt();
		adj 	= new ArrayList[n];
		Root 	= new ArrayList<Integer>();
		for (int i = 0; i < n; ++i) adj[i] = new ArrayList<Integer>();
		for (int i = 0; i < n; ++i) {
			int r = inp.nextInt() - 1;
			if (r == -1) Root.add(i); else adj[r].add(i);
		}
		
		h 		= new int[n];
		p 		= new int[n][20];
		timeIn	= new int[n];
		timeOut	= new int[n];
		Vertex	= new ArrayList[n];
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < 20; ++j) p[i][j] = -1;
			Vertex[i] = new ArrayList<Integer>();
		}
		for (int s: Root) DFS(s);
		Init();
		
		q = inp.nextInt();
		while (q --> 0) {
			int v = inp.nextInt() - 1;
			int p = inp.nextInt();
			int u = KthParent(v, p);
			if (u == -1) out.print("0 ");
			else out.print((Query(u, p) - 1) + " ");
		}
	}
	
	private void DFS(int u) {
		timeIn[u] = cnt++; Vertex[h[u]].add(timeIn[u]);
		for (int v: adj[u]) {
			p[v][0] = u;
			h[v] = h[u] + 1;
			DFS(v);
		}
		timeOut[u] = cnt - 1;
	}
	
	private void Init() {
		for (int j = 1; 1 << j <= n; ++j)
			for (int i = 0; i < n; ++i)
				if (p[i][j - 1] != -1) p[i][j] = p[p[i][j - 1]][j - 1];
	}
	
	private int KthParent(int u, int k) {
		for (int i = 20; i >= 0; --i)
			if ((k >> i & 1) == 1) {
				u = p[u][i];
				if (u == -1) return -1;
			}
		return u;
	}
	
	private int Count(List<Integer> L, int k) {
		int l = 0, r = L.size() - 1, res = 0;
		while (l <= r) {
			int m = (l + r) >> 1;
			if (L.get(m) <= k) {
				res = m + 1;
				l = m + 1;
			}
			else r = m - 1;
		}
		return res;
	}
	
	private int Query(int u, int k) {
		k += h[u];
		return Count(Vertex[k], timeOut[u]) - Count(Vertex[k], timeIn[u] - 1);
	}

}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

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

}