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.ArrayList;
import java.util.StringTokenizer;


public class Main {
    static final int log = 18;
    static ArrayList<Integer> adjList[], queries[];
    static int[] L, P[], queriesValue, ans, size, biggest, cnt, tin, tout, vertex;
    static int time;

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = sc.nextInt();
        queries = new ArrayList[n];
        adjList = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            adjList[i] = new ArrayList<>();
            queries[i] = new ArrayList<>();
        }
        P = new int[n][log];
        L = new int[n];
        tin = new int[n];
        tout = new int[n];
        vertex = new int[n];
        size = new int[n];
        biggest = new int[n];
        ArrayList<Integer> roots = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int p = sc.nextInt() - 1;
            if (p == -1) roots.add(i);
            else adjList[p].add(i);
        }
        for (int root : roots)
            dfs(root, root, 0);
        int m = sc.nextInt();
        ans = new int[m];
        queriesValue = new int[m];
        for (int i = 0; i < m; i++) {
            int v = sc.nextInt() - 1, p = sc.nextInt();
            int anc = findKthAncesstor(v, p);
            if (anc != -1) queries[anc].add(i);
            queriesValue[i] = L[v];
        }
        cnt = new int[n];
        for (int root : roots)
            solve(root, 0);
        for (int x : ans)
            out.print(x + " ");
        out.flush();
        out.close();
    }

    static void solve(int u, int keep) {
        for (int v : adjList[u])
            if (v != biggest[u] && v != P[u][0]) solve(v, 0);
        if (biggest[u] != -1) solve(biggest[u], 1);
        cnt[L[u]]++;
        for (int v : adjList[u]) {
            if (v == biggest[u] || v == P[u][0]) continue;
            for (int i = tin[v]; i <= tout[v]; i++)
                cnt[L[vertex[i]]]++;
        }
        for (int q : queries[u])
            ans[q] = cnt[queriesValue[q]] - 1;
        if (keep == 0)
            for (int i = tin[u]; i <= tout[u]; i++)
                cnt[L[vertex[i]]]--;
    }

    static void dfs(int u, int p, int level) {
        P[u][0] = p;
        L[u] = level;
        size[u] = 1;
        vertex[time] = u;
        tin[u] = time++;
        for (int i = 1; i < log; i++)
            P[u][i] = P[P[u][i - 1]][i - 1];
        biggest[u] = -1;
        for (int v : adjList[u])
            if (v != p) {
                dfs(v, u, level + 1);
                size[u] += size[v];
                if (biggest[u] == -1 || size[biggest[u]] < size[v]) biggest[u] = v;
            }
        tout[u] = time - 1;
    }

    static int findKthAncesstor(int u, int k) {
        if (k > L[u]) return -1;
        for (int i = log; i >= 0; i--)
            if (1 << i <= k) {
                u = P[u][i];
                k -= 1 << i;
            }
        return u;
    }


    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }


        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

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

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

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

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

        public boolean ready() throws IOException {
            return br.ready();
        }


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

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


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

        public double[] nextDoubleArray(int n) throws IOException {
            double[] ans = new double[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextDouble();
            return ans;
        }

        public short nextShort() throws IOException {
            return Short.parseShort(next());
        }

    }
}