Problem: 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

Solution (in JAVA):

Solution:
import com.sun.org.apache.bcel.internal.generic.FieldGen;
import com.sun.org.apache.xpath.internal.operations.Bool;

import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Template implements Runnable {

    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok = new StringTokenizer("");

    void init() throws FileNotFoundException {
        try {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        } catch (Exception e) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
    }

    class GraphBuilder {
        int n, m;
        int[] x, y;
        int index;
        int[] size;


        GraphBuilder(int n, int m) {
            this.n = n;
            this.m = m;
            x = new int[m];
            y = new int[m];
            size = new int[n];
        }

        void add(int u, int v) {
            x[index] = u;
            y[index] = v;
            size[u]++;
            size[v]++;
            index++;
        }

        int[][] build() {
            int[][] graph = new int[n][];
            for (int i = 0; i < n; i++) {
                graph[i] = new int[size[i]];
            }
            for (int i = index - 1; i >= 0; i--) {
                int u = x[i];
                int v = y[i];
                graph[u][--size[u]] = v;
                graph[v][--size[v]] = u;
            }
            return graph;
        }
    }

    String readString() throws IOException {
        while (!tok.hasMoreTokens()) {
            try {
                tok = new StringTokenizer(in.readLine());
            } catch (Exception e) {
                return null;
            }
        }
        return tok.nextToken();
    }

    int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    int[] readIntArray(int size) throws IOException {
        int[] res = new int[size];
        for (int i = 0; i < size; i++) {
            res[i] = readInt();
        }
        return res;
    }

    long[] readLongArray(int size) throws IOException {
        long[] res = new long[size];
        for (int i = 0; i < size; i++) {
            res[i] = readLong();
        }
        return res;
    }

    long readLong() throws IOException {
        return Long.parseLong(readString());
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }

    <T> List<T>[] createGraphList(int size) {
        List<T>[] list = new List[size];
        for (int i = 0; i < size; i++) {
            list[i] = new ArrayList<>();
        }
        return list;
    }

    public static void main(String[] args) {
        new Template().run();
        // new Thread(null, new Template(), "", 1l * 200 * 1024 * 1024).start();
    }

    long timeBegin, timeEnd;

    void time() {
        timeEnd = System.currentTimeMillis();
        System.err.println("Time = " + (timeEnd - timeBegin));
    }

    long memoryTotal, memoryFree;

    void memory() {
        memoryFree = Runtime.getRuntime().freeMemory();
        System.err.println("Memory = " + ((memoryTotal - memoryFree) >> 10)
                + " KB");
    }

    public void run() {
        try {
            timeBegin = System.currentTimeMillis();
            memoryTotal = Runtime.getRuntime().freeMemory();
            init();
            solve();
            out.close();
            if (System.getProperty("ONLINE_JUDGE") == null) {
                time();
                memory();
            }
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }

    class Query {
        int depth;
        int index;

        public Query(int depth, int index) {
            this.depth = depth;
            this.index = index;
        }
    }

    int[] answer;
    int[] name;
    List<Integer>[] graph;
    List<Query>[] queries;

    Set<Integer> merge(Set<Integer> a, Set<Integer> b) {
        if (a.size() > b.size()) {
            a.addAll(b);
            return a;
        } else {
            b.addAll(a);
            return b;
        }
    }

    ArrayList<Set<Integer>> merge(ArrayList<Set<Integer>> a, ArrayList<Set<Integer>> b) {
        if (a.size() < b.size()) {
            return merge(b, a);
        }
        int count = b.size();
        for (int i = 0; i < count; i++) {
            a.set(a.size() - 1 - i, merge(a.get(a.size() - 1 - i), b.get(b.size() - 1 - i)));
        }
        return a;
    }

    ArrayList<Set<Integer>> dfs(int x) {
        ArrayList<Set<Integer>> result = new ArrayList<>();
        for (int y : graph[x]) {
            ArrayList<Set<Integer>> childAns = dfs(y);
            result = merge(result, childAns);
        }
        HashSet<Integer> set = new HashSet<>();
        set.add(name[x]);
        result.add(set);

        for (Query query : queries[x]) {
            int setIndex = result.size() - 1 - query.depth;
            if (setIndex >= 0) {
                answer[query.index] = result.get(setIndex).size();
            }
        }
        return result;
    }

    void solve() throws IOException {
        int n = readInt();
        this.name = new int[n];
        graph = createGraphList(n);
        queries = createGraphList(n);
        HashMap<String, Integer> nameMap = new HashMap<>();
        List<Integer> roots = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            String name = readString();
            int parent = readInt();
            int nameIndex = nameMap.computeIfAbsent(name, k -> nameMap.size());
            this.name[i] = nameIndex;
            if (parent == 0) {
                roots.add(i);
            } else {
                graph[parent - 1].add(i);
            }
        }
        int m = readInt();
        for (int i = 0; i < m; i++) {
            int v = readInt() - 1;
            int k = readInt();
            queries[v].add(new Query(k, i));
        }
        answer = new int[m];
        for (int r : roots) {
            dfs(r);
        }
        for (int ans : answer) {
            out.println(ans);
        }
    }

}