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
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100 * 1000 + 3, MAX_POWER = 10 + 9;
int ancesstor[MAXN][MAX_POWER], height[MAXN], startingTime[MAXN],
    finishingTime[MAXN];
vector<int> adj[MAXN], treeFloor[MAXN];
queue<int> BFSQue;
bool isPassed[MAXN];
void dfs(int root, int parent = -1) {
  isPassed[root] = true;
  ancesstor[root][0] = parent;
  if (ancesstor[root][0] != -1) {
    for (int i = 1; (1 << i) <= height[root]; ++i) {
      ancesstor[root][i] = ancesstor[ancesstor[root][i - 1]][i - 1];
    }
  }
  finishingTime[root] = startingTime[root] + 1;
  for (int i = 0; i < (int)adj[root].size(); ++i) {
    int neighbour = adj[root][i];
    if (!isPassed[neighbour]) {
      startingTime[neighbour] = finishingTime[root];
      height[neighbour] = height[root] + 1;
      treeFloor[height[neighbour]].push_back(startingTime[neighbour]);
      dfs(neighbour, root);
      finishingTime[root] = finishingTime[neighbour] + 1;
    }
  }
  return;
}
int parentFinder(int vertex, int counter) {
  for (int i = 0; (1 << i) <= counter; ++i) {
    if (((counter >> i) & 1) == 1) {
      vertex = ancesstor[vertex][i];
    }
  }
  return vertex;
}
int main() {
  memset(ancesstor, -1, sizeof ancesstor);
  int amountOfVertex;
  cin >> amountOfVertex;
  for (int i = 0; i < amountOfVertex; ++i) {
    int temp;
    cin >> temp;
    temp--;
    if (temp < 0) continue;
    adj[i].push_back(temp);
    adj[temp].push_back(i);
  }
  int startTemp = 0;
  for (int i = 0; i < amountOfVertex; ++i) {
    if (!isPassed[i]) {
      startingTime[i] = startTemp;
      treeFloor[0].push_back(i);
      dfs(i);
      startTemp = finishingTime[i] + 1;
    }
  }
  cerr << "parentFinder(3, 1)"
       << " = " << (parentFinder(3, 1)) << "   ";
  int amountOfQueries;
  cin >> amountOfQueries;
  for (int i = 0; i < amountOfQueries; ++i) {
    int tempV, tempHeight;
    cin >> tempV >> tempHeight;
    tempV--;
    if (height[tempV] < tempHeight) {
      cout << 0 << " " << flush;
      continue;
    }
    int par = parentFinder(tempV, tempHeight);
    int begining =
        upper_bound(treeFloor[height[tempV]].begin(),
                    treeFloor[height[tempV]].end(), startingTime[par]) -
        treeFloor[height[tempV]].begin();
    int ending =
        upper_bound(treeFloor[height[tempV]].begin(),
                    treeFloor[height[tempV]].end(), finishingTime[par]) -
        treeFloor[height[tempV]].begin() - 1;
    cout << (ending - begining) << " " << flush;
  }
  return 0;
}
