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 INF = std::numeric_limits<int>::max();
const char* IN_FILE = ".in";
const char* OUT_FILE = ".out";
int n;
const int NMAX = 100005;
vector<int> g[NMAX];
int nameId[NMAX];
deque<set<int>*>* d[NMAX];
vector<pair<int, int> > q[NMAX];
int ans[NMAX];
void dfs(int v, int p) {
  bool isLeaf = true;
  int maxHeightId = -1;
  for (int i = 0; i < int((g[v]).size()); ++i) {
    int u = g[v][i];
    if (u == p) continue;
    isLeaf = false;
    dfs(u, v);
    if (maxHeightId == -1 || (d[maxHeightId]->size() < d[u]->size())) {
      maxHeightId = u;
    }
  }
  d[v] = new deque<set<int>*>();
  if (isLeaf) {
    d[v]->push_back(new set<int>());
    (*d[v])[0]->insert(nameId[v]);
  } else {
    assert(maxHeightId != -1);
    d[v] = d[maxHeightId];
    deque<set<int>*>& currentD = *d[v];
    for (int i = 0; i < int((g[v]).size()); ++i) {
      int u = g[v][i];
      if (u == p) continue;
      if (u == maxHeightId) continue;
      for (int j = 0; j < int((*d[u]).size()); ++j) {
        if (j >= currentD.size()) {
          assert(false);
        } else {
          if (currentD[j]->size() < (*d[u])[j]->size()) {
            swap(currentD[j], (*d[u])[j]);
          }
          currentD[j]->insert((*d[u])[j]->begin(), (*d[u])[j]->end());
          delete (*d[u])[j];
        }
      }
      d[u]->clear();
    }
    currentD.push_front(new set<int>());
    currentD[0]->insert(nameId[v]);
  }
  for (int it = 0; it < int((q[v]).size()); ++it) {
    pair<int, int> qr = q[v][it];
    if (qr.first < d[v]->size()) {
      ans[qr.second] = (*d[v])[qr.first]->size();
    }
  }
}
void solve() {
  map<string, int> name2Id;
  cin >> n;
  vector<int> roots;
  for (int i = 0; i < int(n); ++i) {
    string name;
    cin >> name;
    if (name2Id.count(name) > 0) {
      nameId[i] = name2Id[name];
    } else {
      nameId[i] = name2Id.size();
      name2Id.insert(make_pair(name, name2Id.size()));
    }
    int p;
    cin >> p;
    --p;
    if (p >= 0) {
      g[p].push_back(i);
    } else
      roots.push_back(i);
  }
  int nq;
  cin >> nq;
  for (int it = 0; it < int(nq); ++it) {
    int v, k;
    cin >> v >> k;
    --v;
    q[v].push_back(make_pair(k, it));
  }
  for (int root : roots) {
    dfs(root, -1);
    for (int i = 0; i < int((*d[root]).size()); ++i) {
      delete (*d[root])[i];
    }
    d[root]->clear();
  }
  for (int i = 0; i < int(nq); ++i) {
    cout << ans[i] << endl;
  }
}
void init() {
  freopen(IN_FILE, "rt", stdin);
  freopen(OUT_FILE, "wt", stdout);
}
int main() {
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}
