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.util.*;
import java.io.*;

public class _0208_E_BloodCousins {

	static ArrayList<Integer> graph[];
	static int st[], et[], arr[], time = 0, dep[], par[], sqrtN;

	public static class Query implements Comparable<Query>{
		int l, r, val, id;
		public Query(int l, int r, int val, int id) {
			this.l = l; this.r = r; this.val = val; this.id = id;
		}
		public int compareTo(Query q) {
			if(this.l/sqrtN == q.l/sqrtN) return Integer.compare(this.r, q.r);
			return Integer.compare(this.l/sqrtN, q.l/sqrtN);

		}
	}

	public static void main(String[] args) throws IOException {
		int N = readInt(); sqrtN = (int) Math.sqrt(N);
		graph = new ArrayList[N+1]; for(int i = 0; i<=N; i++) graph[i] = new ArrayList<>();
		for(int i= 1; i<=N; i++) {int p = readInt(); graph[i].add(p); graph[p].add(i);} 
		st = new int[N+1]; et = new int[N+1]; arr = new int[2*N+1]; dep = new int[N+1]; par = new int[N+1]; dfs(0);
		int logN = Integer.numberOfTrailingZeros(Integer.highestOneBit(N)); int LCA[][] = new int[logN+2][N+1];
		for(int i = 1; i<=N; i++) LCA[0][i] = par[i];
		for(int i = 1; i<=logN+1; i++) for(int j = 1; j<=N; j++) LCA[i][j] = LCA[i-1][LCA[i-1][j]];
		int Q = readInt(); int ans[] = new int[Q]; Query qu[] = new Query[Q]; for(int q = 0; q<Q; q++) {
			int n = readInt(), d = readInt(); if(d >= dep[n]) {ans[q] = 1; qu[q] = new Query(-1, -1, -1, -1);}
			else {
				for(int i = logN+1; i>=0; i--) if(((1<<i)&d) != 0) n = LCA[i][n];
				qu[q] = new Query(st[n], et[n], dep[n]+d, q); 
			}
		}
		Arrays.sort(qu); int l = 1, r = 0, tot[] = new int[N+1]; for(int q = 0; q<Q; q++) {
			int lidx = qu[q].l, ridx = qu[q].r; if(lidx == -1) continue;
			while(l < lidx) tot[arr[l++]]--;
			while(l > lidx) tot[arr[--l]]++;
			while(r < ridx) tot[arr[++r]]++;
			while(r > ridx) tot[arr[r--]]--;
			ans[qu[q].id] += tot[Math.abs(qu[q].val)]/2;
		}
		for(int n : ans) print((n-1) + " "); exit();
	}

	public static void dfs(int n) {
		for(int e : graph[n]) if(par[n] != e){
			par[e] = n; dep[e] = dep[n] + 1; st[e] = ++time; arr[time] = dep[e]; 
			dfs(e); et[e] = ++time; arr[time] = dep[e];
		}
	}

	final private static int BUFFER_SIZE = 1 << 16;
	private static DataInputStream din = new DataInputStream(System.in);
	private static byte[] buffer = new byte[BUFFER_SIZE];
	private static int bufferPointer = 0, bytesRead = 0;
	static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static int readInt() throws IOException {
		int ret = 0;
		byte c = Read();
		while (c <= ' ')
			c = Read();
		do {
			ret = ret * 10 + c - '0';
		} while ((c = Read()) >= '0' && c <= '9');
		return ret;
	}

	private static void fillBuffer() throws IOException {
		bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
		if (bytesRead == -1)
			buffer[0] = -1;
	}

	private static byte Read() throws IOException {
		if (bufferPointer == bytesRead)
			fillBuffer();
		return buffer[bufferPointer++];
	}

	static void print(Object o) {
		pr.print(o);
	}

	static void println(Object o) {
		pr.println(o);
	}

	static void flush() {
		pr.flush();
	}

	static void println() {
		pr.println();
	}

	static void exit() throws IOException {
		din.close();
		pr.close();
		System.exit(0);
	}
}
