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 N = 100005;
vector<int> g[N];
vector<pair<int, int> > query[N];
int name[N], nv[N], sz[N], pesado[N], resp[N];
set<int> vet[N];
void dfs(int u, int p) {
  sz[u] = 1;
  int mx = -1;
  for (int v : g[u]) {
    if (v == p) continue;
    nv[v] = nv[u] + 1;
    dfs(v, u);
    sz[u] += sz[v];
    if (mx == -1 || sz[v] > sz[mx]) mx = v;
  }
  pesado[u] = mx;
}
void add(int u, int p) {
  vet[nv[u]].insert(name[u]);
  for (int v : g[u]) {
    if (v == p) continue;
    add(v, u);
  }
}
void solve(int u, int p, int heavy) {
  for (int v : g[u]) {
    if (v == p || pesado[u] == v) continue;
    solve(v, u, 0);
  }
  if (pesado[u] != -1) solve(pesado[u], u, 1);
  for (int v : g[u]) {
    if (v == p || pesado[u] == v) continue;
    add(v, u);
  }
  vet[nv[u]].insert(name[u]);
  for (int i = 0; i < (int)query[u].size(); i++) {
    int k = query[u][i].first;
    int idx = query[u][i].second;
    resp[idx] = vet[k + nv[u]].size();
  }
  if (heavy == 0) {
    int idx = 0;
    while (vet[idx + nv[u]].size()) {
      vet[idx + nv[u]].clear();
      idx++;
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int cont = 0;
  unordered_map<string, int> mapa;
  vector<int> root;
  for (int i = 1; i <= n; i++) {
    string str;
    cin >> str;
    int x;
    cin >> x;
    if (x != 0)
      g[x].push_back(i), g[i].push_back(x);
    else
      root.push_back(i);
    if (!mapa.count(str)) mapa[str] = cont++;
    name[i] = mapa[str];
  }
  for (int i : root) {
    nv[i] = 1;
    dfs(i, i);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, k;
    cin >> v >> k;
    query[v].push_back(make_pair(k, i));
  }
  for (int i : root) {
    solve(i, i, 0);
  }
  for (int i = 0; i < m; i++) {
    cout << resp[i] << endl;
  }
}
