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 T>
struct fenwick_tree {
  using F = std::function<T(const T &, const T &)>;
  int n;
  std::vector<T> bit;
  F f;
  fenwick_tree() : n(), f() {}
  fenwick_tree(int n, F f = plus<T>()) : n(n), bit(n + 1, 0), f(f) {}
  void add(int i, T val) {
    for (++i; i <= n; i += i & -i) {
      bit[i] = f(bit[i], val);
    }
  }
  T query(int i) {
    T ans = 0;
    for (++i; i; i -= i & -i) {
      ans = f(ans, bit[i]);
    }
    return ans;
  }
};
constexpr int N = 100000 + 1;
int n, a[N], p[N], d[N], timer, tin[N], tout[N];
string name[N];
vector<int> g[N], lvl[N];
vector<tuple<int, int, int>> queries[N];
void dfs(int x) {
  tin[x] = timer++;
  lvl[d[x]].push_back(x);
  for (int i : g[x]) {
    d[i] = 1 + d[x];
    dfs(i);
  }
  tout[x] = timer++;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n, n += 1;
  for (int i = 1; i < n; i++) {
    cin >> name[i] >> p[i];
    g[p[i]].push_back(i);
  }
  dfs(0);
  int mx_depth = *max_element(d, d + n);
  int q;
  cin >> q;
  vector<int> ans(q);
  for (int i = 0, v, k; i < q; i++) {
    cin >> v >> k;
    int dep = d[v] + k;
    if (dep > mx_depth) continue;
    int l = lower_bound(lvl[dep].begin(), lvl[dep].end(), v,
                        [&](int i, int j) { return tin[i] < tin[j]; }) -
            lvl[dep].begin();
    swap(tin[v], tout[v]);
    int r = lower_bound(lvl[dep].begin(), lvl[dep].end(), v,
                        [&](int i, int j) { return tin[i] < tin[j]; }) -
            lvl[dep].begin();
    swap(tin[v], tout[v]);
    if (l < r) {
      queries[dep].emplace_back(l, r - 1, i);
    }
  }
  for (int i = 1; i <= mx_depth; i++) {
    vector<string> names;
    names.reserve(lvl[i].size());
    for (int v : lvl[i]) {
      names.push_back(name[v]);
    }
    sort(names.begin(), names.end());
    names.erase(unique(names.begin(), names.end()), names.end());
    for (int v : lvl[i]) {
      a[v] = lower_bound(names.begin(), names.end(), name[v]) - names.begin();
    }
    fenwick_tree<int> ft(lvl[i].size());
    vector<int> last(names.size(), -1);
    vector<vector<pair<int, int>>> que(lvl[i].size());
    for (auto &[l, r, i] : queries[i]) {
      que[r].emplace_back(l, i);
    }
    for (int r = 0; r < int(lvl[i].size()); r++) {
      int v = lvl[i][r];
      if (last[a[v]] != -1) ft.add(last[a[v]], -1);
      ft.add(last[a[v]] = r, 1);
      for (auto &[l, i] : que[r]) {
        ans[i] = ft.query(r) - ft.query(l - 1);
      }
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}