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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;

public class BloodCousins {


	static int DP[][] = new int[100000+5][61];
	static int in[] = new int[100000+5];
	static int L[] = new int[100000+5];
	static int out[] = new int[100000+5];
	static ArrayList<Integer> roots = new ArrayList();
	static ArrayList<Integer> tree[] = new ArrayList[100000+5];
	static ArrayList<Integer> sub[] = new ArrayList[100000+5];
	static int parent[] = new int[100000+5];
	static int N,q;
	static int time = 0;
	public static void dfs(int idx,int lvl)
	{
		in[idx] = time++;
		L[idx] = lvl;
		sub[lvl].add(in[idx]);
		
		for(Integer k : tree[idx])
		{
			dfs(k,lvl+1);
		}
		out[idx] = time++;
	}

	public static int go(int x,int H)
	{
		int log = 60;
		int exp = L[x] - H;
		if(exp < 0) return 0;
		int lvl = L[x];
		for(int k = log ; k >= 0 ; k--)
		{
			int next = solve(x,k);
			if(L[next] > exp) x = next;	
		}
		x = parent[x];

	
		//System.err.println(x +" "+L[x] +" "+lvl);
		int lo = 0;
		int hi = sub[lvl].size() - 1;
		int lowest = 0;
		int highest = 0;
		while(lo <= hi)
		{
			int mid = lo + (hi - lo) / 2;
			if(sub[lvl].get(mid) > in[x])
			{
				lowest = mid;
				hi = mid - 1;

			}
			else lo = mid + 1;

			
		}


		lo = 0;
		hi = sub[lvl].size() - 1;
		while(lo <= hi)
		{
			int mid = lo + (hi - lo) /2;
			if(sub[lvl].get(mid) < out[x])
			{
				highest = mid;
				lo = mid + 1;

			}
			else hi = mid - 1;
		}



		return Math.max(0,highest - lowest );
	}
	public static int solve(int i,int j)
	{
		if(j == 0) return parent[i];
		if(DP[i][j] != -1) return DP[i][j];
		return DP[i][j] = solve(solve(i,j-1),j-1);
	}
	public static void main(String[]args)throws Throwable
	{
		IR in = new IR(System.in);
		N = in.j();
		//Arrays.fill(parent,-1);
		for(int i = 0 ; i < N ; i++)
		{
			tree[i] = new ArrayList();
			sub[i] = new ArrayList();
		}
		int a = 0;
		
		for(int i = 0 ; i < N ; i++)
		{
			a = in.j() - 1;
			if(a == -1)
			{
				roots.add(i);
				parent[i] = i;
				continue;

			}
			parent[i] = a;
			tree[a].add(i);
		}
		
		

		for(int i = 0 ; i < 100000+5 ; i++) 
			for(int j = 0 ; j < 60 ; j++) DP[i][j] = -1;
		for(Integer k : roots) dfs(k,0);
		q = in.j();
		StringBuilder out = new StringBuilder();
		for(int i = 0 ; i < N ; i++) Collections.sort(sub[i]);
		while(q-- > 0)
		{
			out.append(go(in.j() -1,in.j()) +" ");
		}
		System.out.println(out);
	}



	static	class IR {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public IR(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int j() {
			return Integer.parseInt(next());
		}
		public long ll(){
			return Long.parseLong(next());
		}

	}
}