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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;

public class E246 {

    static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(System.out);
    
    static int nextInt() throws Exception {
        in.nextToken();
        return (int)in.nval;
    }
    
    static String nextString() throws Exception {
        in.nextToken();
        return in.sval;
    }
    
    public static void main(String[] args) throws Exception {
        int n = nextInt();
        Map<String, Integer> names = new HashMap<String, Integer>();
        v = new int[n];
        g = new List[n];
        int[] par = new int[n];
        for (int i = 0; i < n; i++) g[i] = new ArrayList<Integer>();
        for (int i = 0; i < n; i++) {
            String name = nextString();
            if (!names.containsKey(name)) names.put(name, names.size());
            v[i] = names.get(name);
            par[i] = nextInt()-1;
            if (par[i] >= 0) g[par[i]].add(i);
        }
        int m = nextInt();
        ans = new int[m];
        q = new List[n];
        for (int i = 0; i < n; i++) q[i] = new ArrayList<int[]>();
        for (int i = 0; i < m; i++) {
            int a = nextInt() - 1, k = nextInt();
            q[a].add(new int[]{k, i});
        }
        for (int i = 0; i < n; i++) {
            if (par[i] == -1) go(i, -1, 0);
        }
        for (int i = 0; i < m; i++) out.println(ans[i]);
        out.flush();
    }
    
    static int[] ans, v;
    static List<int[]>[] q;
    static List<Integer>[] g;

    static Map<Integer, Set<Integer>> go(int u, int p, int d) {
        Map<Integer, Set<Integer>> map = new HashMap<Integer, Set<Integer>>();
        Set<Integer> d1 = new HashSet<Integer>();
        for (int x : g[u]) if (x != p) {
            Map<Integer, Set<Integer>> cur = go(x, u, d+1);
            map = merge(map, cur);
            d1.add(v[x]);
        }
        map.put(d+1, d1);
        for (int[] qq : q[u]) {
            if (map.containsKey(d+qq[0])) ans[qq[1]] = map.get(d+qq[0]).size();
        }
        return map;
    }
    
    static Set<Integer> merge(Set<Integer> a, Set<Integer> b) {
        if (a == null) return b;
        if (b == null) return a;
        if (a.size() < b.size()) {
            for (int x : a) b.add(x);
            return b;
        }
        else {
            for (int x : b) a.add(x);
            return a;
        }
    }
    
    static Map<Integer, Set<Integer>> merge(Map<Integer, Set<Integer>> a, Map<Integer, Set<Integer>> b) {
        if (a.size() > b.size()) return merge(b, a);
        else {
            for (int x : a.keySet()) {
                b.put(x, merge(a.get(x), b.get(x)));
            }
            return b;
        }
    }
}
