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.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

public class test {
    public static void main(String[] args) {
        new E().run();
    }
}

class E {
    List<Integer>[] a;
    List<int[]>[] q;
    int[] z, r;

    public void run() {
        int n;
        Scanner in = new Scanner(System.in);
        HashMap<String, Integer> g = new HashMap<String, Integer>();
        n = in.nextInt();
        r = new int[n + 1];
        a = new List[n + 1];
        q = new List[n + 1];
        for (int i = 0; i <= n; i++) {
            a[i] = new ArrayList<Integer>();
            q[i] = new ArrayList<int[]>();
        }
        for (int i = 1; i <= n; i++) {
            String s = in.next();
            if (!g.containsKey(s))
                g.put(s, g.size());
            r[i] = g.get(s);
            a[in.nextInt()].add(i);
        }
        int m = in.nextInt();
        z = new int[m];
        for (int i = 0; i < m; i++) {
            int x = in.nextInt(), y = in.nextInt();
            q[x].add(new int[] { y, i });
        }
        dfs(0, 0);
        for (int i = 0; i < m; i++)
            System.out.println(z[i]);
        return;
    }

    Map<Integer, Set<Integer>> dfs(int x, int d) {
        Map<Integer, Set<Integer>> v = new HashMap<Integer, Set<Integer>>(), u;
        Set<Integer> f = new HashSet<Integer>();
        for (int i : a[x]) {
            u = dfs(i, d + 1);
            v = merge(v, u);
            f.add(r[i]);
        }
        v.put(d + 1, f);
        for (int[] i : q[x])
            if (v.containsKey(d + i[0]))
                z[i[1]] = v.get(d + i[0]).size();
        return v;
    }

    Set<Integer> merge(Set<Integer> x, Set<Integer> y) {
        if (x == null)
            return y;
        if (y == null)
            return x;
        if (x.size() < y.size())
            return merge(y, x);
        for (int i : y)
            x.add(i);
        return x;
    }

    Map<Integer, Set<Integer>> merge(Map<Integer, Set<Integer>> x,
            Map<Integer, Set<Integer>> y) {
        if (x.size() < y.size())
            return merge(y, x);
        for (int i : y.keySet())
            x.put(i, merge(x.get(i), y.get(i)));
        return x;
    }
}