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 MAXN = 1000005;
vector<int> ID[MAXN];
vector<int> H[MAXN];
vector<pair<pair<int, int>, int> > Q[MAXN];
vector<int> E[MAXN];
vector<int> roots;
int inicio[MAXN], fin[MAXN];
map<string, int> idx;
bool seen[MAXN];
int height[MAXN];
int TT = 0;
int ans[MAXN];
int valor[MAXN];
int BLOCK_SIZE;
inline bool mo_cmp(const pair<pair<int, int>, int> &x,
                   const pair<pair<int, int>, int> &y) {
  int block_x = x.first.first / BLOCK_SIZE;
  int block_y = y.first.first / BLOCK_SIZE;
  if (block_x != block_y) return block_x < block_y;
  return x.first.second < y.first.second;
}
void dfs(int node, int h) {
  seen[node] = true;
  height[node] = h;
  ID[h].push_back(valor[node]);
  H[h].push_back(TT);
  inicio[node] = TT++;
  for (int i = 0; i < E[node].size(); i++) {
    int to = E[node][i];
    dfs(to, h + 1);
  }
  fin[node] = TT;
}
int cnt[MAXN];
int current_answer = 0;
multiset<int> x;
void add(int valor) {
  if (!x.count(valor)) current_answer++;
  x.insert(valor);
}
void remove(int valor) {
  if (x.count(valor)) x.erase(x.find(valor));
  if (!x.count(valor)) current_answer--;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  BLOCK_SIZE = static_cast<int>(sqrt(n));
  int k = 0;
  for (int i = 1; i <= n; i++) {
    string w;
    int dad;
    cin >> w >> dad;
    if (!dad)
      roots.push_back(i);
    else
      E[dad].push_back(i);
    if (idx.find(w) == idx.end()) idx[w] = k++;
    valor[i] = idx[w];
  }
  for (int i = 1; i <= n; i++) {
    if (!seen[i]) dfs(i, 1);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int v, h;
    cin >> v >> h;
    h += height[v];
    int l = lower_bound(H[h].begin(), H[h].end(), inicio[v]) - H[h].begin();
    int r = lower_bound(H[h].begin(), H[h].end(), fin[v]) - H[h].begin();
    r--;
    if (r < l)
      ans[i] = 0;
    else {
      Q[h].push_back(make_pair(make_pair(l, r), i));
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(Q[i].begin(), Q[i].end(), mo_cmp);
    int mo_left = 0, mo_right = -1;
    for (int j = 0; j < Q[i].size(); j++) {
      int left = Q[i][j].first.first;
      int right = Q[i][j].first.second;
      int id = Q[i][j].second;
      while (mo_right < right) {
        mo_right++;
        add(ID[i][mo_right]);
      }
      while (mo_right > right) {
        remove(ID[i][mo_right]);
        mo_right--;
      }
      while (mo_left < left) {
        remove(ID[i][mo_left]);
        mo_left++;
      }
      while (mo_left > left) {
        mo_left--;
        add(ID[i][mo_left]);
      }
      ans[id] = current_answer;
    }
    x.clear();
    current_answer = 0;
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}
