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;
vector<int> Roots;
string second[200005];
vector<int> adj[200005];
vector<pair<int, string> > P[200005];
vector<pair<pair<int, int>, int> > Q[200005 * 2];
int St[200005];
int En[200005];
int Depth[200005];
int Ans[200005];
map<string, int> CNT;
int Time = 0;
void dfs(int u, int d) {
  St[u] = ++Time;
  Depth[u] = d;
  P[d].push_back(make_pair(Time, second[u]));
  for (int i = 0; i < adj[u].size(); i++) dfs(adj[u][i], d + 1);
  En[u] = ++Time;
}
int main() {
  int n, p;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> second[i];
    scanf("%d", &p);
    adj[p].push_back(i);
    if (p == 0) Roots.push_back(i);
  }
  for (int i = 0; i < n; i++) P[i].push_back(make_pair(INT_MIN, ""));
  for (int i = 0; i < Roots.size(); i++) dfs(Roots[i], 0);
  int q, v, k;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%d %d", &v, &k);
    Q[Depth[v] + k].push_back(make_pair(make_pair(St[v], En[v]), i));
  }
  for (int i = 0; i < n; i++) {
    if (Q[i].size() == 0) continue;
    sort(Q[i].begin(), Q[i].end());
    P[i].push_back(make_pair(INT_MAX, ""));
    CNT.clear();
    int L = 1;
    int R = 0;
    int ans = 0;
    for (int j = 0; j < Q[i].size(); j++) {
      pair<pair<int, int>, int> now = Q[i][j];
      string val;
      while (P[i][L].first > now.first.first) {
        val = P[i][--L].second;
        if (!CNT[val]) ans++;
        CNT[val]++;
      }
      while (P[i][L].first < now.first.first) {
        val = P[i][L++].second;
        CNT[val]--;
        if (!CNT[val]) ans--;
      }
      while (P[i][R].first < now.first.second) {
        val = P[i][++R].second;
        if (!CNT[val]) ans++;
        CNT[val]++;
      }
      while (P[i][R].first > now.first.second) {
        val = P[i][R--].second;
        CNT[val]--;
        if (!CNT[val]) ans--;
      }
      Ans[Q[i][j].second] = ans;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", Ans[i]);
}
