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 class BloodCousins {
	
	public static int N;
	public static int D;
	public static int[][] par;
	public static int[] depth;
	public static boolean[] vis;
	public static ArrayList<Integer>[] adj;
	public static ArrayList<Integer>[] depths;
	public static int[] ti, to;
	public static int time = 0;
	
	public static void main(String[] args) throws Exception{
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		
		N = Integer.parseInt(f.readLine());
		adj = new ArrayList[N];
		depths = new ArrayList[N];
		for(int i = 0; i < N; i ++) {
			adj[i] = new ArrayList<Integer>();
			depths[i] = new ArrayList<Integer>();
		}
		
		D = Integer.numberOfTrailingZeros(Integer.highestOneBit(N));
		par = new int[D+1][N];
		depth = new int[N];
		ti = new int[N];
		to = new int[N];
		
		for(int[] dd:par) {
			Arrays.fill(dd, -1);
		}
		
		StringTokenizer st = new StringTokenizer(f.readLine());
		for(int i = 0; i < N; i ++) {
			int x = Integer.parseInt(st.nextToken()) - 1;
			
			par[0][i] = x;
			if(x != -1) adj[x].add(i);
		}
		
		for(int i = 0; i < N; i ++) {
			if(par[0][i] == -1) {
				dfs(i);
			}
		}
		
		for(int k=1;k<=D;k++) {
			for(int i = 0;i<N;i++) {
				int mid = par[k-1][i];
				if(mid != -1) {
					par[k][i] = par[k-1][mid];
				}
			}
		}
		
		int q = Integer.parseInt(f.readLine());
		for(int i = 0; i < q; i ++) {
			st = new StringTokenizer(f.readLine());
			int x = Integer.parseInt(st.nextToken()) - 1;
			int p = Integer.parseInt(st.nextToken());
			
			System.out.print(solveQuery(x,p) + " ");
		}

	}
	
	static void dfs(int i) {
		ti[i] = time ++;
		for(int x : adj[i]) {
			depth[x] = depth[i] + 1;
			depths[depth[x]].add(x);
			dfs(x);
		}
		to[i] = time ++;
	}
	
	static int walk(int i, int k) {
		for(int d = 0; d <= D && i != -1; d ++) {
			if(((1 << d) & k) > 0) {
				i = par[d][i];
			}
		}
		return i;
	}
	
	static int lca(int i, int j) {
		if(depth[i] > depth[j]) {
			i = walk(i,depth[i] - depth[j]);
		}
		if(depth[j] > depth[i]) {
			j = walk(j,depth[j] - depth[i]);
		}	
		if(i == j) {
			return i;
		}
		
		for(int d=D;d >=0;d--) {
			if(par[d][i] != par[d][j]) {
				i = par[d][i];
				j = par[d][j];
			}
		}
		
		return par[0][i];
	}
	
	static int solveQuery(int i, int p) {
		int x = walk(i,p);
		if(x == -1) return 0;
		
		int left = 0;
		int right = depths[depth[i]].size() - 1;
		while(left < right) {
			int mid = (left + right+1)/2;
			if(ti[depths[depth[i]].get(mid)] < to[x]) {
				left = mid;
			}
			else {
				right = mid - 1;
			}
		}
		
		int rpointer = left;
		
		left = 0;
		right = depths[depth[i]].size() - 1;
		while(left < right) {
			int mid = (left + right)/2;
			if(ti[depths[depth[i]].get(mid)] > ti[x]) {
				right = mid;
			}
			else {
				left = mid + 1;
			}
		}
		
		int lpointer = left;
		
		return rpointer - lpointer;
	}

}