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;
int n, q;
vector<int> root;
vector<pair<int, int> > query[100005];
vector<int> adj[100005];
unordered_map<string, int> mapa;
int v[100005], ans[100005], t;
int pesado[100005], deep[100005], ini[100005], fim[100005], tempo[100005],
    tam[100005];
void dfs(int x) {
  tam[x] = 1;
  tempo[t] = x;
  ini[x] = t++;
  pesado[x] = -1;
  for (int i = 0, y; i < ((int)(adj[x]).size()); i++) {
    y = adj[x][i];
    deep[y] = deep[x] + 1;
    dfs(y);
    tam[x] += tam[y];
    if (pesado[x] == -1 || tam[y] > tam[pesado[x]]) pesado[x] = y;
  }
  fim[x] = t;
}
unordered_set<int> conju[100005 * 2];
unordered_set<int>::iterator it;
void solve(int x, int pai) {
  for (int i = 0, y; i < ((int)(adj[x]).size()); i++) {
    y = adj[x][i];
    if (y == pesado[x]) continue;
    solve(y, x);
  }
  if (pesado[x] != -1) solve(pesado[x], x);
  for (int i = 0, y; i < ((int)(adj[x]).size()); i++) {
    y = adj[x][i];
    if (y == pesado[x]) continue;
    for (int j = ini[y]; j < fim[y]; j++) {
      conju[deep[tempo[j]]].insert(v[tempo[j]]);
    }
  }
  for (int i = 0, h, idx; i < ((int)(query[x]).size()); i++) {
    h = query[x][i].first;
    idx = query[x][i].second;
    ans[idx] = ((int)(conju[deep[x] + h]).size());
  }
  conju[deep[x]].insert(v[x]);
  if (x != pesado[pai]) {
    for (int i = ini[x]; i < fim[x]; i++) {
      it = conju[deep[tempo[i]]].find(v[tempo[i]]);
      if (it != conju[deep[tempo[i]]].end()) conju[deep[tempo[i]]].erase(it);
    }
  }
}
int main() {
  scanf("%d", &n);
  char s[30];
  for (int i = 0, cnt = 0, p; i < n; i++) {
    scanf("%29s", s);
    scanf("%d", &p);
    p--;
    if (p == -1)
      root.push_back(i);
    else
      adj[p].push_back(i);
    if (mapa.find(s) == mapa.end()) mapa[s] = cnt++;
    v[i] = mapa[s];
  }
  scanf("%d", &q);
  for (int i = 0, x, k; i < q; i++) {
    scanf("%d %d", &x, &k);
    x--;
    query[x].push_back(make_pair(k, i));
  }
  pesado[100005 + 1] = -1;
  for (int i = 0; i < ((int)(root).size()); i++) {
    dfs(root[i]);
    solve(root[i], 100005 + 1);
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
}
