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;
struct query {
  int id, dep;
};
const int MAX = 100005;
vector<int> g[MAX];
int depth[MAX];
int sz[MAX];
bool big[MAX];
string name[MAX];
map<string, int> mp[2 * MAX];
vector<query> qry[MAX];
int ans[MAX];
void pre(int u, int dep) {
  depth[u] = dep;
  sz[u] = 1;
  for (auto v : g[u]) {
    pre(v, dep + 1);
    sz[u] += sz[v];
  }
}
void upd(int u, int x) {
  mp[depth[u]][name[u]] += x;
  if (mp[depth[u]][name[u]] == 0) mp[depth[u]].erase(name[u]);
  for (auto v : g[u])
    if (!big[v]) upd(v, x);
}
void dfs(int u, bool keep) {
  int mx = -1, id = -1;
  for (auto v : g[u])
    if (sz[v] > mx) mx = sz[v], id = v;
  for (auto v : g[u])
    if (v != id) dfs(v, false);
  if (id != -1) dfs(id, true), big[id] = true;
  upd(u, 1);
  for (auto x : qry[u]) ans[x.id] = mp[x.dep].size();
  if (id != -1) big[id] = false;
  if (!keep) upd(u, -1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n;
  for (int v = 1, u; v <= n; v++) {
    cin >> name[v] >> u;
    g[u].push_back(v);
  }
  pre(0, 0);
  cin >> q;
  for (int i = 0, u, k; i < q; i++) {
    cin >> u >> k;
    qry[u].push_back({i, depth[u] + k});
  }
  dfs(0, false);
  for (int i = 0; i < q; i++) {
    cout << ans[i] << '\n';
  }
}
