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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.io.StreamTokenizer;
import java.util.LinkedList;


/** Oct 7, 2012 **/

/**
 * @author DOAN Minh Quy
 * @email mquy.doan@gmail.com
 */
public class E208 {

    /**
     * @param args
     * @throws IOException 
     */
    public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub
        new E208().run();
    }
    void run() throws IOException{
        StreamTokenizer token = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintStream printer = new PrintStream(System.out);
        int vertexCount = nextInt(token);
        int[] parentOf = new int[vertexCount];
        @SuppressWarnings("unchecked")
        LinkedList<Integer>[] childList = new LinkedList[vertexCount];
        for(int i = 0 ; i < vertexCount ; ++i){
            childList[i] = new LinkedList<Integer>();
        }
        for(int i = 0 ; i < vertexCount ; ++i){
            parentOf[i] = nextInt(token) - 1;
            if ( parentOf[i] != -1 ){
                childList[parentOf[i]].add(i);
            }
        }
        //---------------------------------------
        int[] order = new int[vertexCount];
        int[] deepth = new int[vertexCount];
        int p = 0;
        for(int i = 0 ; i < vertexCount ; ++i){
            if ( parentOf[i] == -1 ){
                order[p++] = i;
            }
        }
        int prevDeepth = 0;
        int[] firstOfDeepth = new int[vertexCount+1];
        for(int v = 0 ; v < vertexCount ; ++v){
            if ( deepth[order[v]] > prevDeepth ){
                firstOfDeepth[deepth[order[v]]] = v;
                prevDeepth = deepth[order[v]];
            }
            for (int u : childList[order[v]]) {
                order[p++] = u;
                deepth[u] = deepth[order[v]]+1;
            }
        }
        firstOfDeepth[++prevDeepth] = vertexCount;
        int[] inOrder = new int[vertexCount];
        for(int i = 0 ; i < vertexCount ; ++i){
            inOrder[order[i]] = i;
        }
        int[][] sparseTable = sparseParent(parentOf);
        //-------------------------------------------
        int queryCount = nextInt(token);
        for(int query = 1 ; query <= queryCount ; ++query){
            if ( query != 1 ) printer.print(" ");
            int vertex = nextInt(token)-1;
            int distance = nextInt(token);
            int accessor = getAccessor(vertex,distance,sparseTable);
            if( accessor == -1 ){
                printer.print(0);
            }else{
                int left = 0, right = 0;
                {
                    int low = firstOfDeepth[deepth[vertex]]-1;
                    int high = inOrder[vertex];
                    while(high-low>1){
                        int middle = ( high + low ) >> 1;
                        int cousinAccessor = getAccessor(order[middle], distance, sparseTable);
                        if ( cousinAccessor != accessor ){
                            low = middle;
                        }else{
                            high = middle;
                        }
                    }
                    left = high;
                }
                {
                    int low = inOrder[vertex];
                    int high = firstOfDeepth[deepth[vertex]+1];
                    while(high-low>1){
                        int middle = ( high + low ) >> 1;
                        int cousinAccessor = getAccessor(order[middle], distance, sparseTable);
                        if ( cousinAccessor != accessor ){
                            high = middle;
                        }else{
                            low = middle;
                        }
                    }
                    right = high;
                }
                printer.print(right-left-1);
            }
            
        }
    }
    int getAccessor(int vertex,int distance,int[][] sparseTable){
        int accessor = vertex;
        for(int i = 0 ; distance > 0 && accessor != -1  ; distance >>= 1, ++i){
            if ( distance % 2 == 1 ){
                accessor = sparseTable[i][accessor];
            }
        }
        return accessor;
    }
    int[][] sparseParent(int[] parentOf){
        int colCount = parentOf.length;
        int rowCount = Integer.numberOfTrailingZeros(Integer.highestOneBit(colCount-1))+1;
        int[][] sparse = new int[rowCount][colCount];
        sparse[0] = parentOf;
        for(int row = 1 ; row < rowCount ; ++row){
            for(int col = 0 ; col < colCount ; ++col){
                sparse[row][col] = sparse[row-1][col] == -1 ? -1 : sparse[row-1][sparse[row-1][col]];
            }
        }
        return sparse;
    }
    int nextInt(StreamTokenizer token){
        try{
            token.nextToken();
            return (int)token.nval;
        }catch(Exception e){
            return -1;
        }
    }
}