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;
const int N = 1e5 + 7;
int n, m, tot, a[N], L[N], R[N], ans[N], dep[N], last[N];
char str[N];
vector<int> e[N], D[N];
map<string, int> s2i;
struct Query {
  int l, r, d, id;
  bool operator<(const Query &q) const {
    if (d != q.d) return d < q.d;
    return r < q.r;
  }
};
int getid(char *s) {
  if (!s2i.count(s)) {
    int id = ((int)(s2i).size());
    return s2i[s] = id;
  }
  return s2i[s];
}
void dfs(int u, int p) {
  L[u] = ++tot, D[dep[u]].push_back(u);
  for (int i = (0); i < (((int)(e[u]).size())); ++i) {
    int v = e[u][i];
    if (v == p) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
  }
  R[u] = tot;
}
struct Fenwick {
  int n, c[N];
  void init(int _n) {
    n = _n;
    memset(c, 0, sizeof(*c) * (n + 1));
  }
  void upd(int i, int v) {
    while (i <= n) {
      c[i] += v;
      i += i & -i;
    }
  }
  int qry(int i) {
    int r = 0;
    while (i > 0) {
      r += c[i];
      i -= i & -i;
    }
    return r;
  }
} fw;
int main() {
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); ++i) {
    int p;
    scanf(" %s %d", str, &p);
    a[i] = getid(str);
    e[p].push_back(i);
  }
  tot = 0, dfs(0, -1);
  fw.init(n + 1);
  memset(last, -1, sizeof(*last) * (((int)(s2i).size()) + 1));
  vector<Query> Q;
  scanf("%d", &m);
  for (int i = (0); i < (m); ++i) {
    int v, k;
    scanf("%d%d", &v, &k);
    if (dep[v] + k > n) {
      ans[i] = 0;
    } else {
      Q.push_back((Query){L[v], R[v], dep[v] + k, i});
    }
  }
  sort((Q).begin(), (Q).end());
  for (int qi = 0, qj; qi < ((int)(Q).size()); qi = qj) {
    qj = qi + 1;
    while (qj < ((int)(Q).size()) && Q[qj].d == Q[qi].d) ++qj;
    int d = Q[qi].d, di = 0;
    for (int i = (qi); i < (qj); ++i) {
      while (di < ((int)(D[d]).size()) && L[D[d][di]] <= Q[i].r) {
        int u = D[d][di++];
        if (~last[a[u]]) fw.upd(last[a[u]], -1);
        fw.upd(last[a[u]] = L[u], 1);
      }
      ans[Q[i].id] = fw.qry(Q[i].r) - fw.qry(Q[i].l - 1);
    }
    for (int i = (0); i < (di); ++i) {
      int u = D[d][i];
      if (~last[a[u]]) {
        fw.upd(last[a[u]], -1);
        last[a[u]] = -1;
      }
    }
  }
  for (int i = (0); i < (m); ++i) printf("%d\n", ans[i]);
  return 0;
}
