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>
const int DIM = 200005;
int fit[DIM], lat[DIM], ctt[DIM], sz[DIM], hgt[DIM];
std::pair<int, int> qry[DIM];
std::set<int> mst[DIM];
std::map<int, int> mmp[DIM];
std::string str[DIM];
std::vector<int> g[DIM];
std::vector<std::string> lst[DIM];
int dfs1(int x, int h, int l) {
  hgt[x] = h;
  ctt[l] = x;
  fit[x] = lat[x] = l;
  sz[x] = 1;
  for (int y : g[x]) {
    lat[x] = dfs1(y, h + 1, lat[x] + 1);
    sz[x] += sz[y];
  }
  return lat[x];
}
void dfs2(int x, int ok) {
  int mx = -1, p = -1;
  for (int y : g[x]) {
    if (mx < sz[y]) {
      mx = sz[y];
      p = y;
    }
  }
  for (int y : g[x]) {
    if (y != p) {
      dfs2(y, 0);
    }
  }
  if (p != -1) {
    dfs2(p, 1);
  }
  lst[hgt[x]].push_back(str[x]);
  for (int y : g[x]) {
    if (y != p) {
      for (int i = fit[y]; i <= lat[y]; i++) {
        lst[hgt[ctt[i]]].push_back(str[ctt[i]]);
      }
    }
  }
  for (int y : mst[x]) {
    std::sort(lst[y].begin(), lst[y].end());
    int nr = ((lst[y].size() > 0) ? 1 : 0);
    for (int i = 1; i < lst[y].size(); i++) {
      if (lst[y][i] != lst[y][i - 1]) {
        nr++;
      }
    }
    mmp[x][y] = nr;
  }
  if (ok == 0) {
    for (int i = fit[x]; i <= lat[x]; i++) {
      lst[hgt[ctt[i]]].pop_back();
    }
  }
  return;
}
int main(int argc, const char *argv[]) {
  std::ios::sync_with_stdio(false);
  std::cin.tie();
  std::cout.tie();
  int n;
  std::cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    std::cin >> str[i] >> x;
    g[x].push_back(i);
  }
  dfs1(0, 1, 1);
  int m;
  std::cin >> m;
  for (int i = 1; i <= m; i++) {
    std::cin >> qry[i].first >> qry[i].second;
    mst[qry[i].first].insert(qry[i].second + hgt[qry[i].first]);
  }
  dfs2(0, 1);
  for (int i = 1; i <= m; i++) {
    std::cout << mmp[qry[i].first][qry[i].second + hgt[qry[i].first]] << "\n";
  }
  return 0;
}