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 java.util.List;
import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.util.Arrays;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskE {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int log = 1;
        while ((1 << log) < n) ++log;
        int[][] parent = new int[log][n];
        List<Integer>[] children = new List[n];
        for (int i = 0; i < n; i++) {
            children[i] = new ArrayList<Integer>();
        }
        for (int i = 0; i < n; i++) {
            int p = in.nextInt() - 1;
            parent[0][i] = p;
            if (p >= 0) {
                children[p].add(i);
            }
        }
        for (int i = 0; i < n; i++) {
            if (parent[0][i] < 0) {
                for (int j = 0; j < parent.length; j++) {
                    parent[j][i] = -1;
                }
                buildParents(i, parent, children);
            }
        }
        List<Query>[] queries = new List[n];
        int m = in.nextInt();
        for (int i = 0; i < m; i++) {
            int v = in.nextInt() - 1;
            int p = in.nextInt();
            int u = findParent(v, p, parent);
            if (u >= 0) {
                if (queries[u] == null) {
                    queries[u] = new ArrayList<Query>();
                }
                queries[u].add(new Query(i, u, p));
            }
        }
        int[] atDepth = new int[n + 1];
        int[] ans = new int[m];
        Arrays.fill(ans, 1);
        int[] depth = new int[n];
        for (int i = 0; i < n; i++) {
            if (parent[0][i] < 0) {
                dfs(i, queries, children, depth, atDepth, ans);
            }
        }
        for (int i = 0; i < m; i++) {
            out.print((ans[i] - 1) + " ");
        }
    }

    private void dfs(int root, List<Query>[] queries, List<Integer>[] children, int[] depth, int[] atDepth, int[] ans) {
        if (queries[root] != null) {
            for (Query q : queries[root]) {
                ans[q.id] = atDepth[q.depth + depth[root]];
            }
        }
        for (int c : children[root]) {
            depth[c] = 1 + depth[root];
            ++atDepth[depth[c]];
            dfs(c, queries, children, depth, atDepth, ans);
        }
        if (queries[root] != null) {
            for (Query q : queries[root]) {
                ans[q.id] = atDepth[q.depth + depth[root]] - ans[q.id];
            }
        }
    }

    private int findParent(int v, int p, int[][] parent) {
        if (p < 0) {
            return -1;
        }
        if (p == 0) {
            return v;
        }
        for (int i = parent.length - 1; i >= 0; i--) {
            if (parent[i][v] >= 0 && p >= (1 << i)) {
                return findParent(parent[i][v], p - (1 << i), parent);
            }
        }
        return -1;
    }

    private void buildParents(int root, int[][] parent, List<Integer>[] children) {
        for (int c : children[root]) {
            for (int i = 1; i < parent.length; i++) {
                int p = parent[i - 1][c];
                if (p >= 0) {
                    parent[i][c] = parent[i - 1][p];
                } else {
                    parent[i][c] = -1;
                }
            }
            buildParents(c, parent, children);
        }
    }

    class Query {
        int id;
        int v;
        int depth;

        Query(int id, int v, int depth) {
            this.id = id;
            this.v = v;
            this.depth = depth;
        }
    }
}