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 N = 100000;
int n, p[N], m, sz[N], res[N], ans[N];
string s[N];
vector<int> adj[N], q[N], qidx[N];
map<int, set<string> > *mp[N];
void dfs0(int u) {
  sz[u] = 1;
  for (auto v : adj[u]) {
    dfs0(v);
    sz[u] += sz[v];
  }
}
void dfs1(int u) {
  for (auto v : adj[u]) {
    dfs1(v);
  }
  int mx = -1, big_id = -1;
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i];
    if (sz[v] > mx) {
      mx = sz[v];
      big_id = i;
    }
  }
  if (big_id == -1) {
    mp[u] = new map<int, set<string> >();
    (*mp[u])[0].insert(s[u]);
  } else {
    int v = adj[u][big_id];
    mp[u] = mp[v];
    res[u] = res[v] + 1;
    (*mp[u])[0 - res[u]].insert(s[u]);
    for (auto vsmall : adj[u]) {
      if (vsmall != v) {
        for (const auto &el : *mp[vsmall]) {
          (*mp[u])[el.first + res[vsmall] + 1 - res[u]].insert(
              el.second.begin(), el.second.end());
        }
      }
    }
  }
  for (int i = 0; i < q[u].size(); ++i) {
    int k = q[u][i];
    int id = qidx[u][i];
    ans[id] = (*mp[u])[k - res[u]].size();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> s[i] >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; ++i) {
    if (p[i] != -1) {
      adj[p[i]].push_back(i);
    }
  }
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int u, k;
    cin >> u >> k;
    u--;
    q[u].push_back(k);
    qidx[u].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    if (p[i] == -1) {
      dfs0(i);
      dfs1(i);
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << ans[i] << "\n";
  }
}