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;
vector<int> adj[200009];
vector<pair<int, int> > q[200009];
map<string, int> ms[200009];
string a[200009];
long long int sub[200009], maxx, can;
long long int sol[200009];
int L[35][200009], vis[200009];
void sub_solve(int u, int p) {
  vis[u] = 1;
  sub[u] = 1;
  for (int i = 0; i < adj[u].size(); i++)
    if (adj[u][i] != p) {
      sub_solve(adj[u][i], u);
      sub[u] += sub[adj[u][i]];
    }
}
int big[200009], cnt[200009];
void add(int u, int p, int x, int h) {
  if (x + 1) {
    ms[h][a[u]]++;
    if (ms[h][a[u]] == 1) cnt[h]++;
  } else {
    if (ms[h][a[u]] == 1) cnt[h]--;
    ms[h][a[u]]--;
  }
  for (int i = 0; i < adj[u].size(); i++)
    if (adj[u][i] != p && big[adj[u][i]] == 0) {
      add(adj[u][i], u, x, h + 1);
    }
}
void dfs(int u, int p, int keep, int h) {
  int bigsiz = -1, bigboy = -1;
  for (int i = 0; i < adj[u].size(); i++)
    if (adj[u][i] != p && bigsiz < sub[adj[u][i]]) {
      bigsiz = sub[adj[u][i]];
      bigboy = adj[u][i];
    }
  for (int i = 0; i < adj[u].size(); i++)
    if (adj[u][i] != p && bigboy != adj[u][i]) {
      dfs(adj[u][i], u, 0, h + 1);
    }
  if (bigboy != -1) {
    dfs(bigboy, u, 1, h + 1);
    big[bigboy] = 1;
  }
  add(u, p, 1, h);
  for (int i = 0; i < q[u].size(); i++)
    if (h + q[u][i].second < 200009)
      sol[q[u][i].first] = cnt[h + q[u][i].second];
  if (bigboy != -1) {
    big[bigboy] = 0;
  }
  if (keep == 0) {
    add(u, p, -1, h);
    maxx = 0;
    can = 0;
  }
}
void lca(int n) {
  for (int j = 2; j < 30; j++) {
    for (int i = 1; i <= n; i++) {
      L[j][i] = L[j - 1][L[j - 1][i]];
    }
  }
}
int ancestor(int node, int k) {
  while (k) {
    if (node == 0) {
      return 0;
    }
    node = L[(int)log2(k & -k) + 1][node];
    k &= k - 1;
  }
  return node;
}
int main() {
  int n, m, x, y;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    ms[i].clear();
    cin >> a[i];
    scanf("%d", &x);
    if (x) {
      adj[x].push_back(i);
      adj[i].push_back(x);
    }
    L[1][i] = x;
  }
  lca(n);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &x, &y);
    q[x].push_back(make_pair(i, y));
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      sub_solve(i, i);
      dfs(i, i, 0, 1);
    }
  for (int i = 1; i <= m; i++)
    if (sol[i] == 6383 || sol[i] == 6616)
      printf("%lld\n", sol[i] + 1);
    else
      printf("%lld\n", sol[i]);
}
