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;
void read(bool out = 0) {}
vector<vector<int> > ad;
map<string, int> hsh;
int counter = 0;
char input[100];
int a[100009];
bool vis[100009] = {};
int num[100009] = {};
int L[100009] = {};
void dfs(int u, int p) {
  num[u] = 1;
  L[u] = p == -1 ? 1 : L[p] + 1;
  for (auto v : ad[u]) dfs(v, u), num[u] += num[v];
}
set<int> st[100009];
vector<pair<int, int> > q[100009];
int ans[100009];
void add(int u, int big) {
  st[L[u]].insert(a[u]);
  for (auto v : ad[u]) {
    if (v == big) continue;
    add(v, big);
  }
}
void sol(int u) {
  for (int i = 0; i < q[u].size(); i++) {
    ans[q[u][i].second] =
        L[u] + q[u][i].first < 100009 ? st[L[u] + q[u][i].first].size() : 0;
  }
}
void del(int u) {
  if (!st[L[u]].empty()) st[L[u]].clear();
  for (auto v : ad[u]) {
    del(v);
  }
}
void dsu(int u, bool heavy) {
  int mx = -1, big;
  vis[u] = 1;
  for (auto v : ad[u]) {
    if (num[v] > mx) {
      mx = num[v];
      big = v;
    }
  }
  for (auto v : ad[u]) {
    if (v == big) continue;
    dsu(v, 0);
  }
  if (mx != -1) dsu(big, 1);
  add(u, big);
  sol(u);
  if (!heavy) {
    del(u);
  }
}
int main() {
  read();
  int n;
  scanf("%d", &n);
  ad = vector<vector<int> >(n + 1);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%s %d", input, &x);
    string second = string(input);
    if (hsh.count(second))
      a[i] = hsh[second];
    else
      a[i] = hsh[second] = counter++;
    x--;
    if (x != -1) ad[x].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    if (!num[i]) dfs(i, -1);
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d %d ", &x, &y);
    x--;
    q[x].push_back(make_pair(y, i));
  }
  for (int i = 0; i < n; i++) {
    if (!vis[i]) dsu(i, 0);
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}
