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 INF = 0x3f3f3f3f;
const double EPS = 1e-9;
const int R = 100001;
int n, m;
string name[100015];
vector<int> e[100015];
int in[100015], out[100015], dfs_order[100015], cnt_dfs_order, dep[100015];
int siz[100015], son[100015];
set<string> cnt[100015];
void dfs1(int u) {
  dfs_order[++cnt_dfs_order] = u;
  in[u] = cnt_dfs_order;
  siz[u] = 1;
  for (auto v : e[u]) {
    dep[v] = dep[u] + 1;
    dfs1(v);
    if (siz[son[u]] < siz[v]) {
      son[u] = v;
    }
    siz[u] += siz[v];
  }
  out[u] = cnt_dfs_order;
}
int askv[100015], askp[100015];
map<int, int> ask[100015];
int big[100015];
void add(int u, int x) {
  if (1 == x)
    cnt[dep[u]].insert(name[u]);
  else
    cnt[dep[u]].erase(name[u]);
  for (auto v : e[u]) {
    if (!big[v]) {
      add(v, x);
    }
  }
}
void dfs2(int u, int keep) {
  for (auto v : e[u]) {
    if (v == son[u]) continue;
    dfs2(v, 0);
  }
  if (son[u]) {
    dfs2(son[u], 1);
    big[son[u]] = 1;
  }
  add(u, 1);
  for (auto &p : ask[u]) {
    int d = dep[u] + p.first;
    if (d >= R) d = R + 1;
    p.second = cnt[d].size();
  }
  big[son[u]] = 0;
  if (0 == keep) {
    add(u, 0);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> name[i];
    int p;
    cin >> p;
    if (p == 0) p = R;
    e[p].push_back(i);
  }
  dfs1(R);
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> askv[i] >> askp[i];
    ask[askv[i]][askp[i]];
  }
  dfs2(R, 1);
  for (int i = 1; i <= m; ++i) {
    printf("%d\n", ask[askv[i]][askp[i]]);
  }
  return 0;
}
