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 = 100010;
int n, m, t, par, num, Size, Start[N], End[N], Depth[N], bit[N], pr[N];
map<int, int> last, save;
map<string, int> id;
set<int> RowSet;
vector<int> adj[N];
vector<pair<int, int> > Row[N];
struct node {
  int id, l, r;
  bool operator<(const node& cmp) const { return r < cmp.r; }
};
vector<node> req[N];
void Build_Graph() {
  string s;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s >> par;
    if (id[s] == 0) id[s] = ++t;
    save[i] = id[s], adj[par].push_back(i);
  }
}
void DFS_Order(int node, int depth) {
  Start[node] = ++num, Depth[node] = depth;
  Row[Depth[node]].push_back(make_pair(Start[node], node));
  for (int i = 0; i < adj[node].size(); i++) DFS_Order(adj[node][i], depth + 1);
  End[node] = ++num;
}
void Build_Queries() {
  node nd;
  int row, a, b;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    row = Depth[a] + b, nd.id = i;
    nd.l =
        lower_bound(Row[row].begin(), Row[row].end(), make_pair(Start[a], 0)) -
        Row[row].begin();
    nd.r = lower_bound(Row[row].begin(), Row[row].end(), make_pair(End[a], 0)) -
           Row[row].begin();
    if (nd.l == nd.r)
      pr[nd.id] = 0;
    else
      req[row].push_back(nd), RowSet.insert(row);
  }
}
void Add(int ind, int val) {
  while (ind <= Size) {
    bit[ind] += val;
    ind += ind & (-ind);
  }
}
int Query(int ind) {
  int ret = 0;
  while (ind > 0) {
    ret += bit[ind];
    ind -= ind & (-ind);
  }
  return ret;
}
void Solve() {
  for (set<int>::iterator it = RowSet.begin(); it != RowSet.end(); it++) {
    int i = (*it), x = 1;
    Size = Row[i].size();
    last.clear();
    sort(req[i].begin(), req[i].end());
    memset(bit, 0, sizeof bit);
    for (int j = 0; j < req[i].size(); j++) {
      int Id = req[i][j].id, L = req[i][j].l + 1, R = req[i][j].r;
      for (; x <= R; x++) {
        int& val = last[save[Row[i][x - 1].second]];
        if (val != 0) Add(val, -1);
        Add(x, 1), val = x;
      }
      pr[Id] = Query(R) - Query(L - 1);
    }
  }
}
int main() {
  Build_Graph();
  DFS_Order(0, 0);
  Build_Queries();
  Solve();
  for (int i = 0; i < m; i++) cout << pr[i] << "\n";
  return 0;
}
