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.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;



public class Ds {
	
	static int v;
    
	static int[][] par;
	static int[] dist;
	
	static int time = 0;
	
	static int[] start;
	static int[] end;
	
	static List<List<Integer>> adj = new ArrayList<>();
	static 		List<List<Integer>> cum = new ArrayList<>();

	
	public static void main(String[] args) throws NumberFormatException, IOException {

		Scanner in = new Scanner(System.in);
		
		v = in.nextInt();

		for(int i=0; i<=v; i++) {
			adj.add(new ArrayList<>());
			cum.add(new ArrayList<>());
		}
		
		List<Integer> roots = new ArrayList<>();
		
		for(int i=0; i<v; i++) {
			int y = in.nextInt()-1;
			if(y == -1) {
				roots.add(i);
			}
			else {
				adj.get(y).add(i);
				adj.get(i).add(y);
			}
		}
		
		par = new int[21][v];
		dist = new int[v];
		
		start = new int[v];
		end = new int[v];
		
		for(int i=0; i<=20; i++) {
			for(int j=0; j<v; j++) {
				par[i][j] = -1;
			}
		}
		
		for(int i : roots) {
			par[0][i] = -1;
			dfs(i, -1);
		}
		
		int q = in.nextInt();

		for(int i=1; i<=20; i++) {
			for(int j=0; j<v; j++) {
				if(par[i-1][j] != -1) {
					par[i][j] = par[i-1][par[i-1][j]];
				}
			}
		}
		
		PrintWriter out = new PrintWriter(System.out);

		for(int i=0; i<q; i++) {
			int vert = in.nextInt()-1;
			int k = in.nextInt();
			
			if(dist[vert] - k < 0) {
				out.print(0 +" ");
				continue;
			}
			
			int ancestor = findKthAncestor(vert, dist[vert] - k);
			
			int beg = -1;
			int e = -1;
			
			int pos = 0;
			
			for(int j=19; j>=0; j--) {
				int shift = pos + (1 << j);
								
				if(beg + shift < cum.get(dist[vert]).size() && cum.get(dist[vert]).get(beg + shift) < start[ancestor]) {
						beg += shift;
				}
				if(e + shift < cum.get(dist[vert]).size() && cum.get(dist[vert]).get(e + shift) <= end[ancestor]) {
					e += shift;
				}
			}
			
	
			out.print(e - beg - 1 + " ");
			

		}
		
		out.close();
	}


	private static int findKthAncestor(int vert, int k) {
		
		for(int i=20; i>=0; i--) {
			if(par[i][vert] >= 0 && dist[par[i][vert]] >= k) {
				vert = par[i][vert];
			}
		}
		
		return vert;
	}
	
	private static void dfs(int i, int p) {
		start[i] = time;
		cum.get(dist[i]).add(time++);
		
		for(int n : adj.get(i)) {
			if(n == p) {
				continue;
			}
			
			par[0][n] = i;
			dist[n] = dist[i] + 1;
			
			dfs(n, i);
		}
		
		end[i] = time;
	}
}
 