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;
vector<int> child[100007];
unordered_map<string, int> levelName[100007];
int uniqueLevelNames[2 * 100007];
int p[100007];
string name[100007];
int subtreeSize[100007];
int level[100007];
int maxChild[100007];
void dfs(int u) {
  subtreeSize[u] = 1;
  maxChild[u] = -1;
  for (int v : child[u]) {
    level[v] = level[u] + 1;
    dfs(v);
    subtreeSize[u] += subtreeSize[v];
    if (maxChild[u] == -1 || subtreeSize[v] > subtreeSize[maxChild[u]])
      maxChild[u] = v;
  }
}
vector<int> queries[100007];
void add(int u) {
  if (levelName[level[u]][name[u]]++ == 0) uniqueLevelNames[level[u]]++;
  for (int v : child[u]) {
    add(v);
  }
}
void remoove(int u) {
  if (--levelName[level[u]][name[u]] == 0) uniqueLevelNames[level[u]]--;
  for (int v : child[u]) {
    remoove(v);
  }
}
int ans[100007];
void calculate(int u, int keep) {
  for (int v : child[u]) {
    if (v != maxChild[u]) calculate(v, 0);
  }
  if (maxChild[u] != -1) calculate(maxChild[u], 1);
  for (int v : child[u])
    if (v != maxChild[u]) add(v);
  if (levelName[level[u]][name[u]]++ == 0) uniqueLevelNames[level[u]]++;
  for (int i = 0; i < queries[u].size(); i += 2) {
    ans[queries[u][i + 1]] = uniqueLevelNames[level[u] + queries[u][i]];
  }
  if (keep == 0) remoove(u);
}
int main(void) {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> name[i] >> p[i];
    child[p[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (p[i] == 0) dfs(i);
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int v, k;
    cin >> v >> k;
    queries[v].push_back(k);
    queries[v].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (p[i] == 0) calculate(i, 0);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
  return 0;
}
