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
/**
 * DA-IICT
 * Author : Savaliya Sagar
 */
import java.io.*;
import java.math.*;
import java.util.*;

public class Cousins {

	public static int mod = 1000000007;
	public static FasterScanner in = new FasterScanner();
	public static PrintWriter o = new PrintWriter(System.out);

	public static void main(String[] args) {
		new Thread(null, new Runnable() {
			public void run() {
				try {
					Solve();
					o.close();
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
		}, "1", 1 << 26).start();
	}

	public static void Solve() {
		int n = in.nextInt();
		Tree t = new Tree(n);
		int q = in.nextInt();
		while (q-- > 0) {
			int v = in.nextInt();
			int p = in.nextInt();
			o.print(t.lca(v, p) + " ");
		}
	}

	static class Tree {
		ArrayList<ArrayList<Integer>> g = new ArrayList<>();
		int T[]; // parent
		int L[]; // level
		int P[][]; // P[i][j] = 2^j th parent of i
		int time = 0;
		int tin[];
		int tout[];
		int maxH=1;
		ArrayList<Integer> ln[];

		public Tree(int n) {
			T = new int[n + 1];
			L = new int[n + 1];
			P = new int[n + 1][log2(n) + 1];
			tin = new int[n + 1];
			tout = new int[n + 1];
			for (int i = 0; i <= n; i++) {
				g.add(new ArrayList<>());
			}
			for (int i = 1; i <= n; i++) {
				int a = in.nextInt();
				if (a != 0) {
					g.get(a).add(i);
					T[i] = a;
				}
			}
			for (int i = 1; i <= n; i++) {
				if (T[i] == 0) {
					L[i] = 1;
					dfs(i);
				}
			}
			ln = new ArrayList[maxH + 1];
			for (int i = 0; i <= maxH; i++)
				ln[i] = new ArrayList<>();
			for (int i = 1; i <= n; i++) {
				if (T[i] == 0) {
					f(i);
				}
			}
			
			for (int i = 1; i <= n; i++) {
				if (T[i] == 0)
					for (int j = 0; (1 << j) <= n; j++)
						pre(i, j);
			}
		}

		public int log2(int n) {
			return (int) (Math.log10(n) / Math.log10(2));
		}

		public void dfs(int node) {
			time++;
			tin[node] = time;
			for (int k : g.get(node)) {
				L[k] = L[node] + 1;
				if (L[k] > maxH)
					maxH = L[k];
				dfs(k);
			}
			tout[node] = time;
		}

		public void f(int node) {
			ln[L[node]].add(tin[node]);
			for (int k : g.get(node))
				f(k);
		}

		public void pre(int node, int j) {
			if (j == 0) {
				P[node][j] = T[node];
			} else {
				P[node][j] = P[P[node][j - 1]][j - 1];
			}
			for (int k : g.get(node)) {
				pre(k, j);
			}
		}

		public int lca(int v, int p) {
			int temp = v;
			int te = p;
			for (int i = log2(p); i >= 0; i--) {
				if ((1 << i) <= p) {
					temp = P[temp][i];
					p -= (1 << i);
				}
			}
			if(temp==0){
				return 0;
			}

			int s1 = binaryupper(L[temp]+te, tin[temp],0,ln[L[temp]+te].size()-1);
			int s2 = binaryupper(L[temp]+te,tout[temp],0 ,ln[L[temp]+te].size()-1);
			return s2-s1-1;
		}

		public int binaryupper(int id, int m, int l, int r) {

			int mid;
			while (l <= r) {
				mid = (l + r) / 2;
				if (m >= ln[id].get(mid))
					l = mid + 1;
				else
					r = mid - 1;
			}
			return r;
		}

	}

	public static void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	static class FasterScanner {
		private byte[] buf = new byte[1024];
		private int curChar;
		private int snumChars;

		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = System.in.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int[] nextIntArray(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = nextInt();
			}
			return a;
		}

		public long[] nextLongArray(int n) {
			long a[] = new long[n];
			for (int i = 0; i < n; i++) {
				a[i] = nextLong();
			}
			return a;
		}

		private boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
	}
}
