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>
int read() {
  static int ch, x;
  while (isspace(ch = getchar())) {
  }
  x = ch ^ 48;
  while (isdigit(ch = getchar())) x = (((x << 2) + x) << 1) + (ch ^ 48);
  return x;
}
int dfscnt;
int nxt[100010], head[100010], col[100010], cnt[100010], dfn[100010],
    seg[100010], fa[100010], top[100010], siz[100010], hson[100010],
    dep[100010], ans[100010];
std::vector<std::tuple<int, int, int>> vc[100010];
void dfs1(const int &f) {
  siz[f] = 1;
  for (int t = head[f]; t; t = nxt[t]) {
    dep[t] = dep[f] + 1;
    dfs1(t);
    siz[f] += siz[t];
    if (!hson[f] || siz[t] > siz[hson[f]]) hson[f] = t;
  }
}
void dfs2(const int &f) {
  dfn[f] = ++dfscnt;
  seg[dfscnt] = f;
  if (hson[f]) {
    top[hson[f]] = top[f];
    dfs2(hson[f]);
    for (int t = head[f]; t; t = nxt[t]) {
      if (t != hson[f]) {
        top[t] = t;
        dfs2(t);
      }
    }
  }
}
void dfs3(const int &f) {
  ++cnt[dep[f]];
  for (auto &p : vc[f]) std::get<2>(p) = cnt[std::get<0>(p)];
  for (int t = head[f]; t; t = nxt[t]) dfs3(t);
  for (const auto &p : vc[f])
    ans[std::get<1>(p)] = cnt[std::get<0>(p)] - std::get<2>(p) - 1;
}
int anc(int x, int k) {
  while (x) {
    const int delta = dep[x] - dep[top[x]];
    if (k > delta) {
      k -= delta + 1;
      x = fa[top[x]];
    } else {
      x = seg[dfn[x] - k];
      return x;
    }
  }
  return 0;
}
int main() {
  const int n = read();
  for (int i = 1; i <= n; ++i) {
    fa[i] = read();
    nxt[i] = head[fa[i]];
    head[fa[i]] = i;
  }
  dfs1(0);
  dfs2(0);
  const int m = read();
  for (int i = 1; i <= m; ++i) {
    const int x = read(), k = read(), f = anc(x, k);
    if (f) vc[f].push_back(std::make_tuple(dep[x], i, 0));
  }
  dfs3(0);
  std::copy(ans + 1, ans + m + 1, std::ostream_iterator<int>(std::cout, " "));
  return 0;
}
