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 = 200 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
vector<int> a[maxN];
string val[maxN];
int st[maxN], fi[maxN];
int h[maxN], cnt;
int ans[maxN];
vector<pair<int, int> > r[maxN];
vector<pair<pair<int, int>, int> > que[maxN];
map<string, queue<int> > mp;
void dfs(int v, int par = -1) {
  st[v] = cnt++;
  r[h[v]].push_back(make_pair(st[v], v));
  for (auto u : a[v])
    if (u - par) h[u] = h[v] + 1, dfs(u);
  fi[v] = cnt;
}
int fen[maxN], mark[maxN];
void add(int n, int k) {
  for (; n < maxN; n += n & (-n)) fen[n] += k;
}
int get(int n) {
  int res = 0;
  for (; n > 0; n -= n & (-n)) res += fen[n];
  return res;
}
void solve(int t) {
  sort(que[t].begin(), que[t].end());
  int st = 0;
  for (int i = 0; i < r[t].size(); i++) {
    int v = r[t][i].second;
    mp[val[v]].push(i);
  }
  for (int i = 0; i < r[t].size(); i++) {
    int v = r[t][i].second;
    if (mp[val[v]].front() == i) add(i + 1, 1), mark[i] = 1;
  }
  int str = 0;
  for (int i = 0; i < r[t].size(); i++) {
    while (str < que[t].size() && que[t][str].first.first == i) {
      ans[que[t][str].second] = get(que[t][str].first.second);
      str++;
    }
    if (mark[i]) {
      mark[i] = 0;
      add(i + 1, -1);
      int v = r[t][i].second;
      mp[val[v]].pop();
      if (mp[val[v]].size()) {
        mark[mp[val[v]].front()] = 1;
        add(mp[val[v]].front() + 1, 1);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> ro;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> val[i] >> p;
    p--;
    if (p == -1) {
      ro.push_back(i);
      continue;
    }
    a[p].push_back(i);
  }
  for (int i = 0; i < ro.size(); i++) dfs(ro[i]);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int v, le;
    cin >> v >> le;
    v--;
    int tmp = le + h[v];
    int lp = lower_bound(r[tmp].begin(), r[tmp].end(), make_pair(st[v], -1)) -
             r[tmp].begin();
    int rp = lower_bound(r[tmp].begin(), r[tmp].end(), make_pair(fi[v], maxN)) -
             r[tmp].begin();
    que[tmp].push_back(make_pair(make_pair(lp, rp), i));
  }
  for (int i = 0; i < n; i++) solve(i);
  for (int i = 0; i < q; i++) cout << ans[i] << endl;
  return 0;
}
