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
//package codeforces;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class BloodCousins
{
	static int[] ans;
	static ArrayList<query>[] qu;
	public static void main(String[] args) throws IOException
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int[] from = new int[n];
		int[] to = new int[n];
		
		for (int i = 0; i < n; i++)
		{
			from[i] = sc.nextInt()-1;
			to[i] = i;
		}
		
		adj = packD(n, from, to);
		
		int m = sc.nextInt();
		ans = new int[m];
		init(n);
		qu = new ArrayList[n];
		for (int i = 0; i < m; i++)
		{
			int u = sc.nextInt()-1;
			int kth = sc.nextInt();
			int par = -1;
			int up = kth;
			if(lvl[u] >= kth)
			{
				par = u;
				for (int l = lg; l >= 0; l--)
				{
					if(up >= 1<<l)
					{
						up -= 1<<l;
						par = a[l][par];
					}
				}
			}
			if(par == -1)
				ans[i] = 0;
			else
			{
				query q = new query(kth + lvl[par], i);
				if(qu[par] == null)
					qu[par] = new ArrayList<BloodCousins.query>();
				qu[par].add(q);
			}
			
		}
		
		sz = new int[n];
		cnt = new int[n+2];
		for (int i = 0; i < n; i++)
			if(p[i] == -1)
			{
				size(i, -1);
				solve(i, -1, false);
			}
		
		for (int i = 0; i < ans.length; i++)
		{
			if(i != 0)
				pw.print(" ");
			pw.print(ans[i]);
		}
		pw.println();
		pw.flush();
		pw.close();
	}
	
	private static void size(int u, int p)
	{
		sz[u] = 1;
		for (int v : adj[u])
		{
			if(v != p)
			{
				size(v, u);
				sz[u] += sz[v];
			}
		}
	}

	static int[] cnt;
	static void add(int u, int p, int ar, int heavy)
	{
		cnt[lvl[u]]+=ar;
		for (int v : adj[u])
			if(v != p && v != heavy)
				add(v, u, ar, heavy);
	}
	
	static int sz[];
	private static void solve(int u, int p, boolean keep)
	{
		int mx = -1;
		int big = -1;
		for (int v : adj[u])
		{
			if(p == v)
				continue;
			if(sz[v] > mx)
			{
				mx = sz[v];
				big = v;
			}
		}
		
		
		for (int v : adj[u])
			if(v != p && v != big)
				solve(v, u, false);
		
		if(big != -1)
			solve(big, u, true);
		
		add(u, p, 1, big);
		
		if(qu[u] != null)
		for (query q : qu[u])
			ans[q.ind] = cnt[q.l]-1;
		
		if(!keep)
			add(u, p, -1, -1);
	}

	static int[] lvl, p, a[];
	static int lg = 20;
	
	static int[][] adj;
	static int[] q;
	static int s, e;
	static void init(int n)
	{
		lvl = new int[n];
		p = new int[n];
		a = new int[lg+1][n];
		
		q = new int[n];
		Arrays.fill(p, -2);
		for (int i = 0; i < n; i++)
			if(p[i] == -2)
				bfs(i);
		
		for (int i = 0; i < n; i++)
			a[0][i] = p[i];
		
		for (int l = 1; l < lg+1; l++)
			for (int u = 0; u < n; u++)
				if(lvl[u] >= 1<<l)
					a[l][u] = a[l-1][a[l-1][u]];
	}
	
	private static void bfs(int root)
	{
		q[e++] = root;
		p[root] = -1;
		lvl[root] = 0;
		
		for (; s < e;)
		{
			int u = q[s++];
			for (int v : adj[u])
				if(p[u] != v)
				{
					p[v] = u;
					lvl[v] = lvl[u]+1;
					q[e++] = v;
				}
		}
	}
	
	static int[][] packD(int n, int[] from, int[] to) {
		int[][] g = new int[n][];
		int[] p = new int[n];
		for (int f : from) if(f != -1) p[f]++;
		for (int i = 0; i < n; i++) g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++) if(from[i] != -1) {g[from[i]][--p[from[i]]] = to[i];}
		return g;
	}
	
	static class query
	{
		int l, ind;
		public query(int l, int ind)
		{
			this.l = l;
			this.ind = ind;
		}
		
		public String toString()
		{
			return ind + ": " + l;
		}
	}
	static class Scanner
	{
		StringTokenizer st;BufferedReader br;
		public Scanner(InputStream s){br = new BufferedReader(new InputStreamReader(s));}
		public Scanner(String s) throws FileNotFoundException{br = new BufferedReader(new FileReader(new File((s))));}
		public String next() throws IOException{while (st == null || !st.hasMoreTokens())st = new StringTokenizer(br.readLine());return st.nextToken();}
		public int nextInt() throws IOException{return Integer.parseInt(next());}
		public long nextLong() throws IOException{return Long.parseLong(next());}
		public String nextLine() throws IOException{return br.readLine();}
		public double nextDouble() throws IOException{return Double.parseDouble(next());}
		public boolean ready() throws IOException{return br.ready();}
	}
}
