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.*;
import static java.lang.Math.*;

import java.lang.reflect.Array;
import java.util.*;
import java.util.function.*;
import java.lang.*;

public class Main {
    final static boolean debug = false;
    final static String fileName = "";
    final static boolean useFiles = false;

    public static void main(String[] args) throws FileNotFoundException {
        long start;
        if (debug)
            start = System.nanoTime();
        InputStream inputStream;
        OutputStream outputStream;
        if (useFiles) {
            inputStream = new FileInputStream(fileName + ".in");
            outputStream = new FileOutputStream(fileName + ".out");
        } else {
            inputStream = System.in;
            outputStream = System.out;
        }
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task(in, out);
        solver.solve();
        if(debug)
            out.println((System.nanoTime() - start) / 1e+9);
        out.close();
    }
}

class Query {
    int Vertex, Depth, Id;

    Query(int v, int d, int id) {
        Vertex = v;
        Depth = d;
        Id = id;
    }
}

class A extends ArrayList<Integer> {
    @Override
    public Integer get(int i) {
        return super.get(size() - i - 1);
    }

    @Override
    public Integer set(int i, Integer value) {
        return super.set(size() - i - 1, value);
    }
}

class Task {
    int logN = 20;

    A dfs(ArrayList<Integer>[] graph, int[] results, ArrayList<Query>[] queries, int vertex) {
        A x = new A();
        for (int to : graph[vertex]) {
            A y = dfs(graph, results, queries, to);
            if (x.size() < y.size()) {
                A t = x;
                x = y;
                y = t;
            }
            for (int i = 0; i < y.size(); i++) {
                x.set(i, x.get(i) + y.get(i));
            }
        }
        x.add(1);
        for (Query q : queries[vertex]) {
            results[q.Id] = x.get(q.Depth) - 1;
        }
        return x;
    }

    void fillParents(ArrayList<Integer>[] graph, int[][] p, int vertex) {
        for (int j = 1; j < logN; j++) {
            p[vertex][j] = p[p[vertex][j - 1]][j - 1];
        }
        for (int to : graph[vertex])
            fillParents(graph, p, to);
    }

    public void solve() {
        int n = in.nextInt();
        int[][] p = new int[n + 1][logN];
        ArrayList<Integer>[] graph = new ArrayList[n + 1];
        for (int i = 0; i <= n; i++)
            graph[i] = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            p[i][0] = in.nextInt();
            graph[p[i][0]].add(i);
        }

        fillParents(graph, p, 0);

        int m = in.nextInt();
        int[] result = new int[m];
        ArrayList<Query>[] queries = new ArrayList[n + 1];
        for (int i = 0; i <= n; i++)
            queries[i] = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            int v = in.nextInt();
            int depth = in.nextInt();
            int d = depth;
            for (int j = logN - 1; j >= 0; j--) {
                if ((d & (1 << j)) > 0) {
                    d -= 1 << j;
                    v = p[v][j];
                }
            }
            if (v != 0)
                queries[v].add(new Query(v, depth, i));
        }
        dfs(graph, result, queries, 0);
        for (int i = 0; i < m; i++) {
            out.print(result[i] + " ");
        }
    }

    private InputReader in;
    private PrintWriter out;

    Task(InputReader in, PrintWriter out) {
        this.in = in;
        this.out = out;
    }
}

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 double nextDouble(){
        return Double.parseDouble(next());
    }

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

    public char[] nextCharArray(){
        return next().toCharArray();
    }

    public int[] nextIntArray(int n){
        int[] result = new int[n];
        for (int i = 0; i < n; i++)
            result[i] = nextInt();
        return result;
    }

    public long[] nextLongArray(int n){
        long[] result = new long[n];
        for (int i = 0; i < n; i++)
            result[i] = nextLong();
        return result;
    }

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

    public byte nextByte(){
        return Byte.parseByte(next());
    }
}
