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.*;
import static java.lang.Math.*;

public class E {
    public static void main(String[] args) throws Exception {
        InputStream inputStream;
        OutputStream outputStream;
        boolean useFile = false;
        if(useFile == true && System.getProperty("ONLINE_JUDGE") == null) {
            File input = new File("test.inp");
            inputStream = new FileInputStream(input);
        }
        else {
            inputStream = System.in;
        } 
        //inputStream = System.in;
        outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }

    static class Task {

        Vertex[] gr;
        TreeMap < String,Integer > map = new TreeMap < > ();
        TreeMap < Integer,Integer > [] cnt_col;
        ArrayList < Pair > [] queries;
        int[] col;
        int[] cnt_child;
        int[] bigchild;
        int[] st;
        int[] ft;
        int[] ver;
        int[] ans;
        int[] h;
        int n,m;
        int countDiff = 0;
        
        
        public void solve(InputReader in, PrintWriter out) {
            n = in.nextInt();
            gr = new Vertex[n + 2];
            col = new int[n + 2];
            cnt_child = new int[n + 2];
            Arrays.fill(cnt_child,1);
            bigchild = new int[n + 2];
            Arrays.fill(bigchild,-1);
            st = new int[n + 2];
            ft = new int[n + 2];
            ver = new int[n + 2];
            cnt_col = new TreeMap[n + 2];
            h = new int[n + 2];
            Arrays.fill(h,0);
            for(int i = 0;i <= n;++i){
                gr[i] = new Vertex();
                cnt_col[i] = new TreeMap < > ();
            }
            int idx = 0;
            for(int r,i = 1;i <= n;++i){
                String name;
                name = in.next();
                r = in.nextInt();
                if(map.containsKey(name)){
                    col[i] = map.get(name);
                } else {
                    col[i] = idx++;
                    map.put(name,col[i]);
                }
                gr[i].adj.add(r);
                gr[r].adj.add(i);
            }
            m = in.nextInt();
            queries = new ArrayList[n + 2];
            for(int i = 0;i <= n;++i){
                queries[i] = new ArrayList < > ();
            }
            for(int i = 1;i <= m;++i){
                int v,k;
                v = in.nextInt();
                k = in.nextInt();
                queries[v].add(new Pair(k,i));
            }
            preDfs(0,0);
            ans = new int[m + 2];
            dfs(0,0,false);
            for(int i = 1;i <= m;++i){
                out.println(ans[i]);
            }
        }
        
        int TIME = 0;
        void preDfs(int u,int par){
            st[u] = TIME++;
            ver[st[u]] = u;
            for(int v : gr[u].adj){
                if(v != par){
                    h[v] = h[u] + 1;
                    preDfs(v,u);
                    cnt_child[u] += cnt_child[v];
                    if(bigchild[u] == -1 || cnt_child[v] > cnt_child[bigchild[u]]){
                        bigchild[u] = v;
                    }
                }
            }
            ft[u] = TIME;
        }
      
        void add(int f,int val){
            int u = ver[f];
            if(!cnt_col[h[u]].isEmpty() && cnt_col[h[u]].containsKey(col[u])){
                int freq = cnt_col[h[u]].get(col[u]);
                cnt_col[h[u]].remove(col[u]);
                if(freq + val > 0){
                    cnt_col[h[u]].put(col[u],freq + val);
                }
            } else {
                cnt_col[h[u]].put(col[u],1);
            }
        }
        
        void dfs(int u,int par,boolean keep){
            for(int v : gr[u].adj){
                if(v != par && v != bigchild[u]){
                    dfs(v,u,false);
                }
            }
            if(bigchild[u] != -1){
                dfs(bigchild[u],u,true);
            }
            for(int v : gr[u].adj){
                if(v != par && v != bigchild[u]){
                    for(int f = st[v];f < ft[v];++f){
                        add(f,1);
                    }
                }
            }
            try {
                add(st[u],1);
            } catch(Exception e){
                System.out.println("LOZ");
            }
            for(Pair p : queries[u]){
                ans[p.second] = (h[u] + p.first > n ? 0 : cnt_col[h[u] + p.first].size());
            }
            if(keep == false){
                for(int f = st[u];f < ft[u];++f){
                    add(f,-1);
                }
            }
        }
        
        private class Vertex{
            ArrayList < Integer > adj;
            Vertex(){
                adj = new ArrayList < > ();
            }
        }
        
        private class Pair{
            int first;
            int second;
            Pair(int first,int second){
                this.first = first;
                this.second = second;
            }
        }
    }
    
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = reader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
}