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
{
    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 int[] level,parent;
	static ArrayList<Integer>[] al;
	static int[][] st;
	static int LN=21,n;
	static ArrayList<Query> al2[];
	static int[] res;
	static ArrayList<Integer>[] al3;
	static int[] cnt,size;
	
	static void dfs(int u,int p,int curr)
	{
		level[u]=curr;size[u]=1;
		for(int x:al[u])
		{
			if(x!=p)
			{
				dfs(x,u,curr+1);size[u]+=size[x];
			}
		}
	}
	
	static int get(int u,int dis)
	{
		for(int i=LN-1;i>=0;i--)
		{
			int curr=1<<i;
			if(dis>=curr)
			{
				u=st[i][u];dis-=curr;
			}
		}
		return u;
	}
	
	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);cnt[level[y]]++;
				}
			}
		}
		al3[u].add(u);cnt[level[u]]++;
		for(Query x:al2[u])
		{
			if(level[u]+x.val<=n)
			{
				res[x.idx]=cnt[level[u]+x.val];
			}
		}
		if(!keep)
		{
			for(int x:al3[u])
			{
				cnt[level[x]]--;
			}
		}
	}
	
	@SuppressWarnings("unchecked")
    public static void main(String args[]) throws Exception
    {
		n=sc.nextInt();parent=new int[n+1];al=new ArrayList[n+1];al3=new ArrayList[n+1];
		for(int i=1;i<=n;i++)
		{
			parent[i]=sc.nextInt();al[i]=new ArrayList<Integer>();al3[i]=new ArrayList<Integer>();
		}
		cnt=new int[n+1];
		for(int i=1;i<=n;i++)
		{
			if(parent[i]!=0)
			{
				al[parent[i]].add(i);al[i].add(parent[i]);
			}
		}
		level=new int[n+1];size=new int[n+1];st=new int[LN][n+3];
		for(int i=1;i<=n;i++)
		{
			if(parent[i]==0)
			{
				dfs(i,0,1);
			}
		}
		for(int i=1;i<=n;i++)
		{
			st[0][i]=parent[i];
		}
		for(int i=1;i<LN;i++)
		{
			for(int j=1;j<=n;j++)
			{
				st[i][j]=st[i-1][st[i-1][j]];
			}
		}
		al2=new ArrayList[n+1];int m=sc.nextInt();res=new int[m];
		for(int i=1;i<=n;i++)
		{
			al2[i]=new ArrayList<Query>();
		}
		for(int i=0;i<m;i++)
		{
			int idx=i,a=sc.nextInt(),b=sc.nextInt();
			if(level[a]>b)
			{
				int curr=get(a,b);
				al2[curr].add(new Query(i,b));
			}
		}
		for(int i=1;i<=n;i++)
		{
			if(parent[i]==0)
			{
				dfs2(i,0,false);
			}
		}
		for(int i=0;i<m;i++)
		{
			int curr=res[i]==0?0:res[i]-1;
			out.print(curr+" ");
		}
		out.println("");out.close();
    }
}
class Query
{
	int idx,val;
	public Query(int idx,int val)
	{
		this.idx=idx;this.val=val;
	}
}
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());
    }
}