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 N = 100005;
int n, m, name[N], res[N], k[N], par[N];
vector<int> adjList[N], queries[N];
map<int, int> trie[50 * N];
int elem, strcnt, term[50 * N];
int str_hash(string &str) {
  int root = 0;
  for (int c : str) {
    if (!trie[root].count(c)) {
      trie[root][c] = ++elem;
    }
    root = trie[root][c];
  }
  if (!term[root]) {
    term[root] = ++strcnt;
  }
  return term[root];
}
int sz[N], depth[N];
void calcSize(int v) {
  sz[v] = 1;
  for (int u : adjList[v]) {
    depth[u] = depth[v] + 1;
    calcSize(u);
    sz[v] += sz[u];
  }
}
int t, st[N], ft[N], ver[N];
unordered_set<int> nameCnt[N];
void dfs(int v, bool keep) {
  ver[t] = v;
  st[v] = t++;
  int bigChild = 0;
  for (int u : adjList[v]) {
    if (sz[u] > sz[bigChild]) {
      bigChild = u;
    }
  }
  for (int u : adjList[v]) {
    if (u == bigChild) continue;
    dfs(u, 0);
  }
  if (bigChild) {
    dfs(bigChild, 1);
  }
  ft[v] = t;
  for (int u : adjList[v]) {
    if (u == bigChild) continue;
    for (int i = st[u]; i < ft[u]; i++) {
      int w = ver[i];
      nameCnt[depth[w]].insert(name[w]);
    }
  }
  nameCnt[depth[v]].insert(name[v]);
  for (int q : queries[v]) {
    if (k[q] + depth[v] >= N) continue;
    res[q] = nameCnt[k[q] + depth[v]].size();
  }
  if (keep == 0) {
    for (int i = st[v]; i < ft[v]; i++) {
      int u = ver[i];
      nameCnt[depth[u]].erase(name[u]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int v;
  string sname;
  for (int i = 1; i <= n; i++) {
    cin >> sname >> v;
    name[i] = str_hash(sname);
    par[i] = v;
    if (v) {
      adjList[v].push_back(i);
    }
  }
  cin >> m;
  int vi, ki;
  for (int i = 0; i < m; i++) {
    cin >> vi >> ki;
    k[i] = ki;
    queries[vi].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (par[i]) continue;
    calcSize(i);
  }
  for (int i = 1; i <= n; i++) {
    if (par[i]) continue;
    dfs(i, 0);
  }
  for (int i = 0; i < m; i++) {
    cout << res[i] << '\n';
  }
  return 0;
}
