Problem: 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

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
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.StringTokenizer;


public class D {

	static ArrayList<Query>[] queries;
	static ArrayList<Integer>[] next;
	static int[] ans, size, cnt;

	static int findSize(int u)
	{
		int sz = 1;
		for(int v: next[u])
			sz += findSize(v);
		return size[u] = sz;
	}

	static void add(int u, int h, int x)
	{
		cnt[h]++;
		for(int v: next[u])
			if(v != x)
				add(v, h + 1, x);
	}

	static void erase(int u, int h)
	{
		cnt[h]--;
		for(int v: next[u])
			erase(v, h + 1);
	}

	static void dfs(int u, int h, boolean keep)
	{
		int heavy = -1, max = -1;
		for(int v: next[u])
			if(size[v] > max)
				max = size[heavy = v];
		for(int v: next[u])
			if(v != heavy)
				dfs(v, h + 1, false);
		if(heavy != -1)
			dfs(heavy, h + 1, true);
		add(u, h, heavy);

		for(Query q: queries[u])
			ans[q.idx] = cnt[h + q.k] - 1;
		
		if(!keep)
			erase(u, h);
	}


	public static void main(String[] args) throws IOException 
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int N = sc.nextInt();
		int[][] P = new int[18][N];
		for(int i = 0; i < N; ++i)
			P[0][i] = sc.nextInt() - 1;
		for(int k = 1; k < 18; ++k)
			for(int i = 0; i < N; ++i)
				P[k][i] = P[k - 1][i] != -1 ? P[k - 1][P[k - 1][i]] : -1;
				
		int Q = sc.nextInt();
		queries = new ArrayList[N];
		next = new ArrayList[N];
		for(int i = 0; i < N; ++i)
		{
			queries[i] = new ArrayList<Query>();
			next[i] = new ArrayList<Integer>();
		}

		for(int i = 0; i < N; ++i)
			if(P[0][i] != -1)
				next[P[0][i]].add(i);
		for(int i = 0; i < Q; ++i)
		{
			int v = sc.nextInt() - 1, p = sc.nextInt();
			v = getKth(P, p, v);
			if(v == -1)
				continue;
			queries[v].add(new Query(i, p));
		}
		ans = new int[Q];
		size = new int[N];
		cnt = new int[N];
		for(int i = 0; i < N; ++i)
			if(P[0][i] == -1)
			{
				findSize(i);
				dfs(i, 0, false);
			}
		StringBuilder sb = new StringBuilder();
		for(int q: ans)
			sb.append(q + " ");
		out.println(sb);
		out.flush();
		out.close();
	}

	static int getKth(int[][] P, int k, int u)
	{
		for(int i = 17; i >= 0; --i)
			if(1<<i <= k && P[i][u] != -1)
			{
				u = P[i][u];
				k ^= 1<<i;
			}
		return k == 0 ? u : -1;
	}

	static class Query { int idx, k; Query(int a, int b) { idx = a; k = b; } }

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public Scanner(FileReader r){	br = new BufferedReader(r);}

		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();}


	}
}