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>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 1e5 + 10;
const int mod = 1e9 + 7;
bool big[N], root[N];
int t[N], sub[N], cnt[N], bigchild[N], ans[N], L[N], n;
string s[N];
vector<map<string, int> > m;
vector<vector<int> > g;
vector<vector<pair<int, int> > > Q;
void dfs0(int v, int p) {
  sub[v] = 1;
  bigchild[v] = -1;
  int Max = -1;
  for (auto u : g[v]) {
    if (u != p) {
      L[u] = L[v] + 1;
      dfs0(u, v);
      sub[v] += sub[u];
      if (sub[u] > Max) {
        Max = sub[u];
        bigchild[v] = u;
      }
    }
  }
  return;
}
void add(int v, int p) {
  m[L[v]][s[v]]++;
  if (m[L[v]][s[v]] == 1) cnt[L[v]]++;
  for (auto u : g[v])
    if (u != p && (!big[u])) add(u, v);
  return;
}
void rem(int v, int p) {
  m[L[v]][s[v]]--;
  if (m[L[v]][s[v]] == 0) cnt[L[v]]--;
  for (auto u : g[v])
    if ((u != p) && (!big[u])) rem(u, v);
  return;
}
void dfs2(int v, int p, int keep) {
  int i;
  for (auto u : g[v])
    if (u != p && u != bigchild[v]) dfs2(u, v, false);
  if (bigchild[v] != -1) {
    dfs2(bigchild[v], v, true);
    big[bigchild[v]] = true;
  }
  add(v, p);
  for (auto x : Q[v]) {
    if (x.first + L[v] <= n) {
      t[x.second] = cnt[x.first + L[v]];
    } else
      t[x.second] = 0;
  }
  big[bigchild[v]] = false;
  if (keep == false) rem(v, p);
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int i, x, q;
  cin >> n;
  g.resize(n + 2);
  Q.resize(n + 2);
  m.resize(n + 2);
  for (i = 1; i < n + 1; i++) {
    cin >> s[i] >> x;
    if (x != 0) {
      g[x].push_back(i);
      g[i].push_back(x);
    } else
      root[i] = true;
  }
  int z;
  cin >> z;
  for (i = 1; i < z + 1; i++) {
    int a, b;
    cin >> a >> b;
    Q[a].push_back({b, i});
  }
  for (i = 1; i < n + 1; i++) {
    if (root[i]) {
      dfs0(i, i);
      dfs2(i, i, false);
    }
  }
  for (i = 1; i < z + 1; i++) cout << t[i] << endl;
  return 0;
}