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 CPP):

Solution:
#include <bits/stdc++.h>
int N, M;
int men[100000];
struct cquery {
  int v, k;
  int res;
} queries[100000];
std::vector<int> children[100000];
std::vector<int> roots;
std::vector<int> dfsqueries[100000];
void readinput() {
  std::cin >> N;
  std::map<std::string, int> names;
  for (int i = 0; i < N; i++) {
    std::string name;
    int parent;
    std::cin >> name >> parent;
    if (!names.count(name)) {
      int size = names.size();
      names[name] = size;
    }
    men[i] = names[name];
    if (parent > 0) {
      parent--;
      children[parent].push_back(i);
    } else {
      roots.push_back(i);
    }
  }
  std::cin >> M;
  for (int i = 0; i < M; i++) {
    std::cin >> queries[i].v >> queries[i].k;
    queries[i].v--;
    dfsqueries[queries[i].v].push_back(i);
  }
}
int size[100000];
std::set<int> bydepth[100000];
void dfs_size(int node) {
  size[node] = 1;
  for (int i = 0; i < children[node].size(); i++) {
    int child = children[node][i];
    dfs_size(child);
    size[node] += size[child];
  }
}
void dfs_ins(int node, int depth) {
  for (int i = 0; i < children[node].size(); i++) {
    int child = children[node][i];
    dfs_ins(child, depth + 1);
  }
  bydepth[depth].insert(men[node]);
}
void dfs_clr(int node, int depth) {
  for (int i = 0; i < children[node].size(); i++) {
    int child = children[node][i];
    dfs_clr(child, depth + 1);
  }
  bydepth[depth].erase(men[node]);
}
void dfs_calc(int node, int depth) {
  std::pair<int, int> heavy(0, -1);
  for (int i = 0; i < children[node].size(); i++) {
    int child = children[node][i];
    heavy = std::max(heavy, std::make_pair(size[child], child));
  }
  for (int i = 0; i < children[node].size(); i++) {
    int child = children[node][i];
    if (child != heavy.second) {
      dfs_calc(child, depth + 1);
      dfs_clr(child, depth + 1);
    }
  }
  if (heavy.second >= 0) {
    dfs_calc(heavy.second, depth + 1);
  }
  for (int i = 0; i < children[node].size(); i++) {
    int child = children[node][i];
    if (child != heavy.second) {
      dfs_ins(child, depth + 1);
    }
  }
  bydepth[depth].insert(men[node]);
  for (int i = 0; i < dfsqueries[node].size(); i++) {
    struct cquery& q = queries[dfsqueries[node][i]];
    if (depth + q.k >= 100000) {
      q.res = 0;
    } else {
      q.res = bydepth[depth + q.k].size();
    }
  }
}
int main() {
  readinput();
  for (int i = 0; i < roots.size(); i++) {
    dfs_size(roots[i]);
  }
  for (int i = 0; i < roots.size(); i++) {
    dfs_calc(roots[i], 0);
    dfs_clr(roots[i], 0);
  }
  for (int i = 0; i < M; i++) {
    std::cout << queries[i].res << std::endl;
  }
  return 0;
}