Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
map<string, int> mapper;
int N, M, color[maxn], sz[maxn], big[maxn], tmp[2 * maxn], depth[maxn];
unordered_map<int, int> cnt[maxn];
unordered_map<int, int> res[maxn];
unordered_set<int> depths[maxn];
vector<int> graph[maxn];
void addd(int c, int d) {
  if (++cnt[d][c] == 1) {
    tmp[d]++;
  }
}
void remove(int c, int d) {
  if (--cnt[d][c] == 0) {
    tmp[d]--;
  }
}
void init(int u, int p, int d) {
  sz[u] = 1;
  depth[u] = d;
  for (auto &v : graph[u]) {
    if (v == p) continue;
    init(v, u, d + 1);
    sz[u] += sz[v];
  }
}
void add(int u, int p, int x) {
  if (x == 1)
    addd(color[u], depth[u]);
  else
    remove(color[u], depth[u]);
  for (auto &v : graph[u]) {
    if (v == p || big[v]) continue;
    add(v, u, x);
  }
}
void dfs(int u, int p, int keep) {
  int maxi = -1, idx = -1;
  for (auto &v : graph[u]) {
    if (v == p) continue;
    if (sz[v] > maxi) {
      maxi = sz[v];
      idx = v;
    }
  }
  for (auto &v : graph[u]) {
    if (v == p || v == idx) continue;
    dfs(v, u, 0);
  }
  if (idx != -1) {
    dfs(idx, u, 1);
    big[idx] = 1;
  }
  add(u, p, 1);
  for (auto &x : depths[u]) {
    res[u][x] = tmp[x + depth[u]];
  }
  if (idx != -1) big[idx] = 0;
  if (!keep) add(u, p, -1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  vector<int> root;
  int c = 1;
  for (int i = 0; i < N; i++) {
    string str;
    int p;
    cin >> str >> p;
    p--;
    if (mapper.find(str) == mapper.end()) {
      mapper[str] = c++;
    }
    color[i] = mapper[str];
    if (p == -1) {
      root.push_back(i);
      continue;
    }
    graph[i].push_back(p);
    graph[p].push_back(i);
  }
  cin >> M;
  vector<pair<int, int> > queries;
  while (M--) {
    int u, d;
    cin >> u >> d;
    u--;
    depths[u].insert(d);
    queries.push_back({u, d});
  }
  for (auto &x : root) {
    init(x, x, 0);
    dfs(x, x, 0);
  }
  for (auto &x : queries) {
    cout << res[x.first][x.second] << '\n';
  }
  return 0;
}