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

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author George Marcus
 */
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 {
    private int N;
    private int M;
    private int idx;
    private int[] T;
    private int[] L;
    private int[] number;
    private int[] cnt;
    private int[][] dp;
    private ArrayList<Integer>[] graph;
    //private SortedSet<Integer>[] H;
    private ArrayList<Integer>[] H;
    private static final int LOGN = 20;

    public void solve(int testNumber, Scanner in, PrintWriter out) {
        N = in.nextInt();
        T = new int[N + 1];
        L = new int[N + 1];
        number = new int[N + 1];
        cnt = new int[N + 1];
        graph = new ArrayList[N + 1];
        for(int i = 1; i <= N; i++)
            graph[i] = new ArrayList<Integer>();
        int t;
        for(int i = 1; i <= N; i++) {
            t = in.nextInt();
            T[i] = t;
            if(t > 0)
                graph[t].add(i);
        }
        M = in.nextInt();

        //H = new SortedSet[N + 1];
        H = new ArrayList[N + 1];
        for(int i = 0; i <= N; i++)
            //H[i] = new TreeSet<Integer>();
            H[i] = new ArrayList<Integer>();
        idx = 0;
        for(int i = 1; i <= N; i++)
            if(T[i] == 0)
                dfs(i, 0);

        dp = new int[N + 1][LOGN + 1];
        for(int i = 1; i <= N; i++)
            dp[i][0] = T[i];
        for(int j = 1; (1 << j) <= N; j++) // invers?
            for(int i = 1; i <= N; i++)
                dp[i][j] = dp[dp[i][j - 1]][j - 1];

        int v, p, u;
        for(int i = 0; i < M; i++) {
            v = in.nextInt();
            p = in.nextInt();
            u = go(v, p);

            if(u == 0) {
                out.print("0 ");
                continue;
            }
            //int res = H[L[v]].subSet(number[u], number[u] + cnt[u]).size() - 1;
            int res, min, max;
            min = Collections.binarySearch(H[L[v]], number[u]);
            min = -min - 1;
            max = Collections.binarySearch(H[L[v]], number[u] + cnt[u] - 1);
            res = max >= 0 ? max - min : -max - 1 - min - 1;

            //res = Math.max(res, 0);

            out.print(res + " ");
        }
	}

    private int dfs(int node, int level) {
        L[node] = level;
        number[node] = idx++;
        H[level].add(number[node]);
        cnt[node] = 1;

        for(int i = 0; i < graph[node].size(); i++)
            cnt[node] += dfs(graph[node].get(i), level + 1);

        return cnt[node];
    }

    private int go(int node, int dist) {
        int bit = 0;
        while(dist > 0) {
            if((dist & 1) > 0) node = dp[node][bit];
            bit++;
            dist /= 2;
        }
        return node;
    }
}

