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 maxn = 1e5 + 10;
vector<int> grafo[maxn];
int n, m, sz[maxn], level[maxn], big[maxn], ans[maxn];
string cor[maxn];
map<string, int> freq[maxn];
vector<pair<int, int> > q[maxn];
void dfs(int u) {
  sz[u] = 1;
  for (auto v : grafo[u]) {
    level[v] = level[u] + 1;
    dfs(v);
    sz[u] += sz[v];
  }
}
void add(int u, int x) {
  freq[level[u]][cor[u]] += x;
  if (!freq[level[u]][cor[u]]) freq[level[u]].erase(cor[u]);
  for (auto v : grafo[u])
    if (!big[v]) add(v, x);
}
void sack(int u, bool keep) {
  int idmax = 0;
  for (auto v : grafo[u])
    if (sz[idmax] < sz[v]) idmax = v;
  for (auto v : grafo[u])
    if (v != idmax) sack(v, 0);
  if (idmax) sack(idmax, 1), big[idmax] = 1;
  add(u, 1);
  for (auto x : q[u])
    if (level[u] + x.second < maxn)
      ans[x.first] = (int)freq[level[u] + x.second].size();
  big[idmax] = 0;
  if (!keep) add(u, -1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> cor[i] >> p;
    grafo[p].push_back(i);
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int v, k;
    cin >> v >> k;
    q[v].push_back({i, k});
  }
  for (int i = 1; i <= n; i++) {
    if (!sz[i]) {
      dfs(i);
      sack(i, 0);
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
  return 0;
}
