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;
map<string, int> No;
set<int> si[100005];
string str;
int col[100005], dep[100005], hson[100005], size[100005];
vector<int> tre[100005];
void init(int u) {
  size[u] = 1;
  for (int i = 0; i < tre[u].size(); i++) {
    int v = tre[u][i];
    dep[v] = dep[u] + 1;
    init(v);
    if (size[v] > size[hson[u]]) hson[u] = v;
    size[u] += size[v];
  }
}
int HS;
void add(int u, bool flag) {
  if (flag)
    si[dep[u]].insert(col[u]);
  else
    si[dep[u]].erase(col[u]);
  for (int i = 0; i < tre[u].size(); i++) {
    int v = tre[u][i];
    if (v == HS) continue;
    add(v, flag);
  }
}
struct Ques {
  int id, k;
};
vector<Ques> vq[100005];
int ans[100005];
bool vis[100005];
void dfs(int u, bool kep) {
  vis[u] = true;
  for (int i = 0; i < tre[u].size(); i++) {
    int v = tre[u][i];
    if (v != hson[u]) dfs(v, false);
  }
  if (hson[u]) {
    dfs(hson[u], true);
    HS = hson[u];
  }
  add(u, true);
  HS = 0;
  for (int i = 0; i < vq[u].size(); i++) {
    Ques q = vq[u][i];
    ans[q.id] = si[q.k].size();
  }
  if (!kep) add(u, false);
}
bool isson[100005];
int main() {
  int n, cnt = 0, rot;
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1, f; i <= n; i++) {
    cin >> str >> f;
    if (No[str] == 0) No[str] = ++cnt;
    col[i] = No[str];
    if (f) {
      tre[f].push_back(i);
      isson[i] = true;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!isson[i]) {
      dep[i] = 1;
      init(i);
    }
  int m;
  cin >> m;
  for (int i = 1, v, d; i <= m; i++) {
    cin >> v >> d;
    if (d + dep[v] > n) continue;
    vq[v].push_back((Ques){i, d + dep[v]});
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, false);
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  return 0;
}
