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.*;
import java.util.*;
public final class blood_cousins_return
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	static TreeMap<String,Integer>[] tm;
	static int[][] al;
	static String[] a;
	static int[] parent,level,size,res;
	static List<Node> list=new ArrayList<Node>();
	static ArrayList<Node>[] al2;
	static ArrayList<Integer>[] al3;
	
	static void dfs1(int u,int p,int curr)
	{
		level[u]=curr;size[u]=1;
		for(int x:al[u])
		{
			if(x!=p)
			{
				dfs1(x,u,curr+1);size[u]+=size[x];
			}
		}
	}
	
	static void dfs2(int u,int p,boolean keep)
	{
		int max=-1,big_child=-1;
		for(int x:al[u])
		{
			if(x!=p && size[x]>max)
			{
				max=size[x];big_child=x;
			}
		}
		for(int x:al[u])
		{
			if(x!=p && x!=big_child)
			{
				dfs2(x,u,false);
			}
		}
		if(big_child!=-1)
		{
			dfs2(big_child,u,true);al3[u]=al3[big_child];
		}
		for(int x:al[u])
		{
			if(x!=p && x!=big_child)
			{
				for(int y:al3[x])
				{
					al3[u].add(y);put(level[y],a[y],1);
				}
			}
		}
		put(level[u],a[u],1);al3[u].add(u);
		for(Node x:al2[u])
		{
			res[x.u]=tm[x.v].size();
		}
		if(!keep)
		{
			for(int x:al3[u])
			{
				put(level[x],a[x],-1);
			}
		}
	}
	
	static void put(int idx,String curr,int val)
	{
		if(tm[idx].get(curr)==null)
		{
			tm[idx].put(curr,val);
		}
		else
		{
			tm[idx].put(curr,tm[idx].get(curr)+val);
		}
		if(tm[idx].get(curr)==0)
		{
			tm[idx].remove(curr);
		}
	}
	
	@SuppressWarnings("unchecked")
    public static void main(String args[]) throws Exception
    {
		int n=sc.nextInt();a=new String[n];parent=new int[n];int[] count=new int[n];
		for(int i=0;i<n;i++)
		{
			parent[i]=-1;a[i]=sc.next();int u=i,v=sc.nextInt()-1;
			if(v>=0)
			{
				list.add(new Node(u,v));count[u]++;count[v]++;parent[u]=v;
			}
		}
		al=new int[n][];al2=new ArrayList[n];al3=new ArrayList[n];tm=new TreeMap[n];
		for(int i=0;i<n;i++)
		{
			al[i]=new int[count[i]];al2[i]=new ArrayList<Node>();al3[i]=new ArrayList<Integer>();tm[i]=new TreeMap<>();
		}
		count=new int[n];level=new int[n];size=new int[n];
		for(int i=0;i<list.size();i++)
		{
			int u=list.get(i).u,v=list.get(i).v;
			al[u][count[u]]=v;al[v][count[v]]=u;count[u]++;count[v]++;
		}
		for(int i=0;i<n;i++)
		{
			if(parent[i]==-1)
			{
				dfs1(i,-1,0);
			}
		}
		int q=sc.nextInt();res=new int[q];
		for(int i=0;i<q;i++)
		{
			int u=sc.nextInt()-1,k=sc.nextInt();
			if(level[u]+k<n)
			{
				al2[u].add(new Node(i,level[u]+k));
			}
		}
		for(int i=0;i<n;i++)
		{
			if(parent[i]==-1)
			{
				dfs2(i,-1,false);
			}
		}
		for(int i=0;i<q;i++)
		{
			out.println(res[i]);
		}
		out.close();
    }
}
class Node
{
	int u,v;
	public Node(int u,int v)
	{
		this.u=u;this.v=v;
	}
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}