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>
void sci(T& t) {
  cin >> t;
}
template <typename T, typename... Ts>
void sci(T& t, Ts&... ts) {
  sci(t);
  sci(ts...);
}
string nm[111111];
vector<int> g[111111];
bool is_root[111111];
int tin[111111];
int tout[111111];
vector<int> lvl[111111];
int h[111111];
struct query {
  int l, r, id;
  bool operator<(const query& q) const { return r < q.r; }
};
vector<query> qs[111111];
int timer = 0;
void dfs(int v, int p, int d) {
  tin[v] = ++timer;
  lvl[d].push_back(v);
  h[v] = d;
  for (int i : g[v]) {
    if (i != p) {
      dfs(i, v, d + 1);
    }
  }
  tout[v] = ++timer;
}
int ans[111111];
struct SegmentTree {
  int* a;
  int n;
  SegmentTree() {}
  SegmentTree(int n) : n(n) {
    a = new int[4 * n];
    memset(a, 0, sizeof(int) * 4 * n);
  }
  void set(int i, int l, int r, int qi, int qx) {
    if (l + 1 == r) {
      a[i] = qx;
      return;
    }
    int m = (l + r) >> 1;
    if (qi < m) {
      set(i * 2 + 1, l, m, qi, qx);
    } else {
      set(i * 2 + 2, m, r, qi, qx);
    }
    a[i] = a[i * 2 + 1] + a[i * 2 + 2];
  }
  void set(int i, int x) { set(0, 0, n, i, x); }
  int get(int i, int l, int r, int ql, int qr) {
    if (r <= ql || qr <= l) {
      return 0;
    }
    if (ql <= l && r <= qr) {
      return a[i];
    }
    int m = (l + r) >> 1;
    return get(i * 2 + 1, l, m, ql, qr) + get(i * 2 + 2, m, r, ql, qr);
  }
  int get(int l, int r) { return get(0, 0, n, l, r); }
};
int last[111111];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  sci(n);
  for (int i = 0; i < n; i++) {
    string s;
    sci(s);
    int x;
    sci(x);
    --x;
    nm[i] = s;
    if (x != -1) {
      g[x].push_back(i);
      g[i].push_back(x);
    } else {
      is_root[i] = true;
    }
  }
  for (int i = 0; i < n; i++) {
    if (is_root[i]) {
      dfs(i, -1, 0);
    }
  }
  int m;
  sci(m);
  for (int i = 0; i < m; i++) {
    int x, y;
    sci(x, y);
    --x;
    y += h[x];
    if (y >= n || lvl[y].empty()) {
      continue;
    }
    auto cmp_tin = [&](const int& a, const int& b) { return tin[a] < tin[b]; };
    auto cmp_tout = [&](const int& a, const int& b) {
      return tout[a] < tout[b];
    };
    int l =
        upper_bound(lvl[y].begin(), lvl[y].end(), x, cmp_tin) - lvl[y].begin();
    int r =
        upper_bound(lvl[y].begin(), lvl[y].end(), x, cmp_tout) - lvl[y].begin();
    qs[y].push_back(query{l, r, i});
  }
  SegmentTree st(n);
  map<string, int> ids;
  for (int i = 0; i < n; i++) {
    sort(qs[i].begin(), qs[i].end());
    memset(st.a, 0, sizeof(int) * 4 * lvl[i].size());
    st.n = lvl[i].size();
    memset(last, -1, sizeof(int) * lvl[i].size());
    ids.clear();
    int lr = 0;
    for (auto& q : qs[i]) {
      if (q.r != lr) {
        for (int j = lr; j < q.r; j++) {
          int t = lvl[i][j];
          if (ids.find(nm[t]) == ids.end()) {
            int id = ids.size();
            ids[nm[t]] = id;
          }
          t = ids[nm[t]];
          if (last[t] != -1) {
            st.set(last[t], 0);
          }
          st.set(j, 1);
          last[t] = j;
        }
      }
      lr = q.r;
      ans[q.id] = st.get(q.l, q.r);
    }
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}
