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 java.util.*;

public class R151D2E {

	Scanner in=new Scanner(System.in);
	PrintWriter out=new PrintWriter(System.out);

	int[] id,pa,ret;
	List<Integer>[] tr;
	List<int[]>[] q;

	Map<Integer, Set<Integer>> gao(int w,int p,int l) {
		Map<Integer, Set<Integer>> mp=new TreeMap<Integer, Set<Integer>>();
		Set<Integer> st=new HashSet<Integer>();
		st.add(id[w]);
		for(int i: tr[w]){
			if(i==p) continue;
			mp=merge(mp,gao(i,w,l+1));
		}
		mp.put(l,st);
		for(int[] i: q[w])
			if(mp.containsKey(i[0]+l))
				ret[i[1]]=mp.get(i[0]+l).size();
		return mp;
	}

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

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

	void run() {
		int n=in.nextInt(),m;
		id=new int[n];
		pa=new int[n];
		tr=new List[n];
		Map<String,Integer> name=new HashMap<String,Integer>();
		for(int i=0; i<n; i++)
			tr[i]=new ArrayList<Integer>();
		for(int i=0; i<n; i++) {
			String s=in.next();
			int x=in.nextInt();
			if(!name.containsKey(s))
				name.put(s,name.size());
			id[i]=name.get(s);
			pa[i]=x-1;
			if(pa[i]>=0)
				tr[pa[i]].add(i);
		}

		m=in.nextInt();
		ret=new int[m];
		q=new List[n];
		for(int i=0; i<n; i++)
			q[i]=new ArrayList<int[]>();
		for(int i=0; i<m; i++) {
			int x=in.nextInt(),y=in.nextInt();
			q[x-1].add(new int[]{y,i});
		}

		for(int i=0; i<n; i++)
			if(pa[i]==-1)
				gao(i,-1,0);
		for(int i=0; i<m; i++)
			out.println(ret[i]);
		out.flush();
	}

	static public void main(String[] args) throws Exception{
		new R151D2E().run();
	}
}
