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;
using ll = long long;
string to_string(const string s) { return '"' + s + '"'; }
string to_string(const char c) { return char(39) + string(1, c) + char(39); }
string to_string(const char* s) { return to_string(string(s)); }
string to_string(bool f) { return (f ? "true" : "false"); }
template <class A, class B>
string to_string(const pair<A, B> x) {
  return "(" + to_string(x.first) + ", " + to_string(x.second) + ")";
}
template <class A, class B, class C>
string to_string(const tuple<A, B, C> x) {
  return "(" + to_string(get<0>(x)) + ", " + to_string(get<1>(x)) + ", " +
         to_string(get<2>(x)) + ")";
}
template <class A, class B, class C, class D>
string to_string(const tuple<A, B, C, D> x) {
  return "(" + to_string(get<0>(x)) + ", " + to_string(get<1>(x)) + ", " +
         to_string(get<2>(x)) + ", " + to_string(get<3>(x)) + ")";
}
template <class T>
string to_string(const T v) {
  string res = "{";
  bool f = true;
  for (auto x : v) res += (f ? to_string(x) : ", " + to_string(x)), f = false;
  return res + "}";
}
void debug_args() { cerr << "]\n"; }
template <class H, class... T>
void debug_args(H x, T... y) {
  cerr << to_string(x);
  if (sizeof...(y)) cerr << ", ";
  debug_args(y...);
}
const int N = 3e5 + 13;
int n, m, ans[N], dp[N], depth[N];
string str[N];
map<string, int> mp[N];
vector<int> g[N], root;
vector<pair<int, int>> q[N];
void dfsSz(int v, int d = 1) {
  dp[v] = 1;
  depth[v] = d;
  for (int to : g[v]) {
    dfsSz(to, d + 1);
    dp[v] += dp[to];
  }
}
void add(int v) { mp[depth[v]][str[v]]++; }
void del(int v) {
  mp[depth[v]][str[v]]--;
  if (mp[depth[v]][str[v]] == 0) mp[depth[v]].erase(str[v]);
}
void go(int v, int t) {
  if (t)
    del(v);
  else
    add(v);
  for (int to : g[v]) go(to, t);
}
void dfs(int v, bool keep) {
  int node = -1;
  for (int to : g[v]) {
    if (node == -1 || dp[node] < dp[to]) node = to;
  }
  for (int to : g[v]) {
    if (to != node) dfs(to, 0);
  }
  if (node != -1) dfs(node, -1);
  add(v);
  for (int to : g[v]) {
    if (to != node) go(to, 0);
  }
  for (auto it : q[v]) {
    ans[it.second] = mp[it.first + depth[v]].size();
  }
  if (!keep) go(v, 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int v = 1; v <= n; v++) {
    string s;
    int u;
    cin >> s >> u;
    if (u)
      g[u].push_back(v);
    else
      root.push_back(v);
    str[v] = s;
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, k;
    cin >> v >> k;
    q[v].push_back(make_pair(k, i));
  }
  for (int v : root) {
    dfsSz(v);
    dfs(v, 0);
  }
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
  return 0;
}