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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.Collections;

public class E {

	static StreamTokenizer st;
	static ArrayList<Integer>[]ages;
	static int[][]up;
	static int L, timer;
	static int[]tin,tout,height;
	public static void main(String[] args) throws IOException{
		st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = nextInt();
		int[]fath = new int[n+1];
		ages = new ArrayList[n+1];
		for (int i = 1; i <= n; i++) {
			ages[i] = new ArrayList<Integer>();
		}
		for (int i = 1; i <= n; i++) {
			fath[i] = nextInt();
			if (fath[i] != 0)
				ages[fath[i]].add(i);
		}
		while ((1 << L) <= n)
			L++;
		up = new int[n+1][L+1];
		tin = new int[n+1];
		tout = new int[n+1];
		height = new int[n+1];
		height[0] = -1;
		for (int i = 1; i <= n; i++) {
			if (fath[i]==0)
				dfs(i, 0);
		}
		ArrayList<Integer>[]H = new ArrayList[n+1];
		for (int i = 0; i <= n; i++) {
			H[i] = new ArrayList<Integer>();
		}
		for (int i = 1; i <= n; i++) {
			H[height[i]].add(tin[i]);
		}
		for (int i = 0; i <= n; i++) {
			Collections.sort(H[i]);
		}
		int m = nextInt();
		for (int i = 1; i <= m; i++) {
			int v = nextInt();
			int p = nextInt();
			int pth_fath = v;
			for (int j = L; j >= 0; j--) {
				if ((1 << j) < p) {
					pth_fath = up[pth_fath][j];
					p -= (1 << j);
				}
			}
			pth_fath = up[pth_fath][0];
			if (pth_fath==0)
				pw.print(0+" ");
			else {
				int h = height[v];
				int ind1 = Collections.binarySearch(H[h], tin[pth_fath]);
				int ind2 = Collections.binarySearch(H[h], tout[pth_fath]);
				if (ind1 < 0)
					ind1 = Math.abs(ind1);
				if (ind2 < 0)
					ind2 = Math.abs(ind2)-1;
				int ans = ind2-ind1+1;
				ans--;
				pw.println(ans+" ");
			}
		}
		pw.close();
	}
	
	private static void dfs(int v, int p) {
		tin[v] = timer++;
		height[v] = height[p]+1;
		up[v][0] = p;
		for (int i = 1; i <= L; i++) {
			up[v][i] = up[up[v][i-1]][i-1];
		}
		for (int to : ages[v]) {
			if (to != p)
				dfs(to, v);
		}
		tout[v] = timer++;
	}

	private static int nextInt() throws IOException{
		st.nextToken();
		return (int) st.nval;
	}
}