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 CF130E {
    public static PrintWriter out;
    public static FastScanner in;

    private ArrayList<Integer>[] G;
    private ArrayList<Integer>[] D;
    private int timer;
    private int[] inTime;
    private int[] outTime;
    private int[][] parent;
    private int[] level;
    private void solve() {
        int n = in.nextInt();

        G = new ArrayList[n+1];
        for (int i=0;i<=n;i++){
            G[i] = new ArrayList<>();
        }

        List<Integer> roots = new ArrayList<>();
        for(int i=1;i<=n;i++){
            int u = in.nextInt();
            if(u > 0){
                G[u].add(i);
            }else{
                roots.add(i);
            }
        }
        timer = 0;
        inTime = new int[n+1];
        outTime = new int[n+1];
        parent = new int[22][n+1];
        level = new int[n+1];
        D = new ArrayList[n+1];
        for(int i=0;i<=n;i++){
            D[i] = new ArrayList<>();
        }

        for (int i:roots){
            DFS(i,0,0);
        }
        int q = in.nextInt();
        for(int i=0;i<q;i++){
            int v = in.nextInt();
            int p = in.nextInt();

            if(level[v] < p){
                out.print("0 ");
            }else{
                int l = level[v];
                for(int j=20;j>=0;j--){
                    if(l - level[parent[j][v]] < p){
                        v = parent[j][v];
                    }
                }
                v = parent[0][v];
                int x = Collections.binarySearch(D[l],inTime[v]);
                int y = Collections.binarySearch(D[l],outTime[v]);

                if(x < 0){
                    x = -(x+1);
                }
                if(y < 0){
                    y = -(y+1);
                }

                out.print((y-x-1)+" ");
            }
        }

    }

    private void DFS(int u, int p, int depth){
        level[u] = depth;
        inTime[u] = ++timer;
        D[level[u]].add(timer);
        parent[0][u] = p;

        for(int j=1;j<21;j++){
            parent[j][u] = parent[j-1][parent[j-1][u]];
        }

        for (int v:G[u]){
            DFS(v,u,depth+1);
        }
        outTime[u] = ++timer;
    }

    private void runIO() {
        in = new FastScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        solve();
        out.close();
    }

    private static class FastScanner {
        BufferedReader bufferedReader;
        StringTokenizer stringTokenizer;

        private FastScanner() {
            bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        }

        private String next() {
            if (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
                try {
                    stringTokenizer = new StringTokenizer(bufferedReader.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return stringTokenizer.nextToken();
        }

        private int nextInt() {
            return Integer.parseInt(next());
        }

        private long nextLong() {
            return Long.parseLong(next());
        }

        private String nextLine() {
            String ret = "";
            try {
                ret = bufferedReader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return ret;
        }
    }

    public static void main(String[] args) {
        new CF130E().runIO();
    }
}