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;
int n, m;
vector<int> e[100010];
map<string, int> idx;
int val[100010];
string s;
int cnt;
int a;
int d[100010];
int last[100010];
int w[100010];
int w2[100010];
vector<int> nodes[100010];
int dfs(int p, int de) {
  d[p] = de;
  w[p] = cnt++;
  w2[cnt - 1] = p;
  last[p] = cnt - 1;
  for (__typeof((e[p]).begin()) it = (e[p]).begin(); it != (e[p]).end(); it++) {
    last[p] = dfs(*it, de + 1);
  }
  return last[p];
}
struct ST {
  int N;
  vector<int> u;
  vector<vector<int> > v;
  ST(int _n, vector<int> _w) {
    N = _n;
    v = vector<vector<int> >(4 * _n, vector<int>(0, 0));
    u = _w;
    update(1, 1, N);
  }
  ST() {}
  void update(int id, int l, int r) {
    if (l == r) {
      v[id].push_back(u[l - 1]);
      return;
    }
    int h = (l + r) / 2;
    update(2 * id, l, h);
    update(2 * id + 1, h + 1, r);
    for (int i = 0; i < v[2 * id].size(); i++) v[id].push_back(v[2 * id][i]);
    for (int i = 0; i < v[2 * id + 1].size(); i++)
      v[id].push_back(v[2 * id + 1][i]);
    sort(v[id].begin(), v[id].end());
  }
  int query(int l, int r, int ma) { return QUERY(1, 1, N, l, r, ma); }
  int QUERY(int id, int l, int r, int lt, int rt, int ma) {
    if (l > rt || r < lt) return 0;
    if (l >= lt && r <= rt) {
      int s = upper_bound(v[id].begin(), v[id].end(), ma) - v[id].begin();
      return (r - l + 1) - s;
    }
    int h = (l + r) / 2;
    int t = 0;
    t += QUERY(2 * id, l, h, lt, rt, ma);
    t += QUERY(2 * id + 1, h + 1, r, lt, rt, ma);
    return t;
  }
};
vector<ST> st;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s >> a;
    if (!idx[s]) idx[s] = ++cnt;
    val[i + 1] = idx[s];
    e[a].push_back(i + 1);
  }
  cnt = 0;
  dfs(0, 0);
  for (int i = 0; i <= n; i++) {
    nodes[d[i]].push_back(w[i]);
  }
  for (int i = 0; i <= n; i++) sort(nodes[i].begin(), nodes[i].end());
  for (int i = 0; i <= n; i++) {
    if (nodes[i].size() == 0) continue;
    vector<int> next;
    for (int j = 0; j < nodes[i].size(); j++) {
      next.push_back(nodes[i].size() + 2);
    }
    vector<pair<int, int> > tmp;
    for (int j = 0; j < nodes[i].size(); j++)
      tmp.push_back(make_pair(val[w2[nodes[i][j]]], j));
    sort(tmp.begin(), tmp.end());
    for (int j = 0; j < tmp.size() - 1; j++) {
      if (tmp[j + 1].first == tmp[j].first) {
        next[tmp[j].second] = tmp[j + 1].second;
      }
    }
    st.push_back(ST(next.size(), next));
  }
  int q;
  cin >> q;
  for (int qq = 0; qq < q; qq++) {
    int v, k;
    scanf("%d%d", &v, &k);
    int dep = d[v] + k;
    if (dep > n) {
      printf("0\n");
      continue;
    }
    int l, r;
    l = lower_bound(nodes[dep].begin(), nodes[dep].end(), w[v]) -
        nodes[dep].begin();
    r = lower_bound(nodes[dep].begin(), nodes[dep].end(), last[v]) -
        nodes[dep].begin();
    if (r >= nodes[dep].size() || nodes[dep][r] > last[v]) r--;
    if (r < l) {
      printf("0\n");
    } else {
      int res = st[dep].query(l + 1, r + 1, r);
      printf("%d\n", res);
    }
  }
  return 0;
}
