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.util.*;
import java.io.*;
import java.math.*;
 
public class Main {
    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(System.out);
        //Scanner sc = new Scanner();
        Reader in = new Reader();
        Main solver = new Main();
        solver.solve(out, in);
        out.flush();
        out.close();
 
    }
    
    static int log2(int a){
        return (int)(Math.log(a)/Math.log(2));
    }

    static int INF = (int)1e9;
    static int maxn = (int)1e5+5;
    static long mod=(int)1e9+7 ;
    static int n,m,k,q,t;
    
    static int[][] par;
    static int[] vin,vout,depth;
    static ArrayList<Integer> adj[],lvl[];

    void solve(PrintWriter out, Reader in) throws IOException{   
        n = in.nextInt();

        vin = new int[n+1];
        vout = new int[n+1];
        depth = new int[n+1];
        adj = new ArrayList[n+1];
        lvl = new ArrayList[n+1];
        for(int i=1;i<=n;i++) {
            adj[i] = new ArrayList<Integer>();
            lvl[i] = new ArrayList<Integer>();
        }
        par = new int[n+1][18];
        for(int i=1;i<=n;i++){
            par[i][0] = in.nextInt();
            if(par[i][0]!=0) adj[par[i][0]].add(i);
        }
        
        for(int i=1;i<18;i++){
            for(int j=1;j<=n;j++){
                par[j][i] = par[par[j][i-1]][i-1];
            }
        }

        for(int i=1;i<=n;i++){
            if(par[i][0]==0) DFS(i,0);
        }

        q = in.nextInt();
        int v,dist=0,log=0,l=0,r=0;
        for(int i=0;i<q;i++){
            v = in.nextInt();
            k = in.nextInt();

            if(depth[v]<=k) out.print(0+" ");
            else{
                dist = k;
                while(dist!=0){
                    log = log2(dist);
                    v = par[v][log];
                    dist -= 1<<log;
                }

                k+=depth[v];

                l = low(vin[v],k);
                r = top(vout[v],k);
                out.print((r-l)+" ");
            }
        }
        out.println();
    }
    
    static int time=0;

    static void DFS(int v,int par){
        vin[v] = time++;
        depth[v] = depth[par]+1;
        lvl[depth[v]].add(v);

        for(int e:adj[v]){
            if(e!=par){
                DFS(e,v);
            }
        }
        vout[v] = time++;
    }

    static int low(int val,int level){
        int lo=0,hi=lvl[level].size()-1,mid,res=-1;

        while(lo<=hi){
            mid = (lo+hi)/2;
            if(vin[lvl[level].get(mid)]>=val){
                hi = mid-1;
                res = mid;
            }else lo = mid+1;
        }
        return res;
    }

    static int top(int val,int level){
        int lo=0,hi=lvl[level].size()-1,mid,res=-1;

        while(lo<=hi){
            mid = (lo+hi)/2;
            if(vin[lvl[level].get(mid)]<=val){
                lo = mid+1;
                res = mid;
            }else hi = mid-1;
        }
        return res;
    }

    static class Reader {

    private InputStream mIs;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;

    public Reader() {
        this(System.in);
    }

    public Reader(InputStream is) {
        mIs = is;
    }

    public int read() {
        if (numChars == -1) {
            throw new InputMismatchException();

    }
        if (curChar >= numChars) {
            curChar = 0;
            try {
                numChars = mIs.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 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 String next() {
        int c = read();
        while (isSpaceChar(c)) {
            c = read();
        }
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }

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

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

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

    }
}