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;
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& _p) {
  return os << "(" << _p.first << "," << _p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& _V) {
  bool f = true;
  os << "[";
  for (auto v : _V) {
    os << (f ? "" : ",") << v;
    f = false;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& _S) {
  bool f = true;
  os << "(";
  for (auto s : _S) {
    os << (f ? "" : ",") << s;
    f = false;
  }
  return os << ")";
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const map<T, U>& _M) {
  return os << set<pair<T, U>>(_M.begin(), _M.end());
}
const signed long long INF = 1000000100;
const long double EPS = 1e-9;
const int MAXN = 500010;
int N, Q;
vector<int> G[MAXN];
char buff[MAXN];
string A[MAXN];
vector<pair<int, int>> queries[MAXN];
void read_data() {
  scanf("%d", &N);
  for (int(i) = (1); (i) <= (N); (i)++) {
    int a;
    scanf("%s %d", buff, &a);
    A[i] = string(buff);
    G[a].push_back(i);
  }
  scanf("%d", &Q);
  for (int(i) = (1); (i) <= (Q); (i)++) {
    int w, h;
    scanf("%d %d", &w, &h);
    queries[w].push_back(make_pair(h, i));
  }
}
int first[MAXN];
int x_v[MAXN];
int S[MAXN];
int H[MAXN];
int ts = 0;
map<string, int> M[MAXN];
int R[MAXN];
inline void modify(int v, int delta) {
  int h = H[v];
  const string& s = A[v];
  M[h][s] += delta;
  if (M[h][s] == 0) M[h].erase(s);
}
void dfs1(int v, int p, int h) {
  first[v] = ++ts;
  x_v[first[v]] = v;
  S[v] = 1;
  H[v] = h;
  for (int u : G[v]) {
    dfs1(u, v, h + 1);
    S[v] += S[u];
  }
}
void dfs2(int v, int p, bool keep) {
  int w = -1;
  int max_s = -1;
  for (int u : G[v])
    if (S[u] > max_s) {
      max_s = S[u];
      w = u;
    }
  for (int u : G[v])
    if (u != w) dfs2(u, v, false);
  if (w != -1) {
    dfs2(w, v, true);
  }
  for (int u : G[v])
    if (u != w)
      for (int(i) = (first[u]); (i) <= (first[u] + S[u] - 1); (i)++) {
        modify(x_v[i], 1);
      }
  modify(v, 1);
  for (auto q : queries[v]) {
    int h = H[v] + q.first;
    int query_id = q.second;
    R[query_id] = int(M[h].size());
  }
  if (!keep)
    for (int(i) = (first[v]); (i) <= (first[v] + S[v] - 1); (i)++)
      modify(x_v[i], -1);
}
void solve() {
  dfs1(0, -1, 1);
  dfs2(0, -1, false);
  for (int(i) = (1); (i) <= (Q); (i)++) printf("%d\n", R[i]);
}
int main() {
  read_data();
  solve();
}
