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 const N = 1e5 + 10;
int n, q, ans[N];
vector<int> adj[N];
vector<pair<int, int> > query[N];
int id[N], cont = 0;
map<string, int> Map;
int sz[N], prof[N];
void dfs(int v) {
  sz[v] = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    int w = adj[v][i];
    prof[w] = prof[v] + 1;
    dfs(w);
    sz[v] += sz[w];
  }
}
int cnt[N], pointer[N];
map<int, int> comp[N];
vector<int> total[N];
bool bad[N];
void add(int v, int x) {
  if (v > 0) {
    int p = prof[v], idx = id[v];
    int pos = comp[p][idx];
    total[p][pos] += x;
    if (x == -1 && total[p][pos] == 0) cnt[p]--;
    if (x == 1 && total[p][pos] == 1) cnt[p]++;
  }
  for (int i = 0; i < adj[v].size(); i++) {
    int w = adj[v][i];
    if (!bad[w]) add(w, x);
  }
}
void solve(int v, bool keep) {
  int heavy = -1;
  for (int i = 0; i < adj[v].size(); i++) {
    int w = adj[v][i];
    if (heavy == -1 || sz[heavy] < sz[w]) heavy = w;
  }
  for (int i = 0; i < adj[v].size(); i++) {
    int w = adj[v][i];
    if (w != heavy) solve(w, 0);
  }
  if (heavy != -1) solve(heavy, 1), bad[heavy] = 1;
  add(v, 1);
  for (int i = 0; i < query[v].size(); i++) {
    int p = query[v][i].first, idx = query[v][i].second;
    ans[idx] = cnt[p];
  }
  if (heavy != -1) bad[heavy] = 0;
  if (!keep) add(v, -1);
}
int main() {
  scanf(" %d ", &n);
  for (int i = 1; i <= n; i++) {
    char s[25];
    int p;
    scanf(" %s %d ", s, &p);
    if (!Map[(string)(s)]) Map[(string)(s)] = ++cont;
    id[i] = Map[(string)(s)];
    adj[p].push_back(i);
  }
  dfs(0);
  for (int i = 1; i <= n; i++) total[i].push_back(0);
  for (int i = 1; i <= n; i++) {
    int p = prof[i], idx = id[i];
    if (!comp[p][idx]) {
      comp[p][idx] = ++pointer[p];
      total[p].push_back(0);
    }
  }
  scanf(" %d", &q);
  for (int i = 1; i <= q; i++) {
    int v, k;
    scanf(" %d %d", &v, &k);
    if (prof[v] + k <= n) query[v].push_back(pair<int, int>(prof[v] + k, i));
  }
  solve(0, 1);
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
