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;
string name[100005];
bool used[100005];
vector<vector<int> > adj;
int nameid[100005];
vector<int> dpth[100005];
int ndpth[100005];
vector<int> accum[100005];
int X;
int L[100005], R[100005];
void dfs(int node, int d) {
  ndpth[node] = d;
  dpth[d].push_back(node);
  L[node] = X;
  ++X;
  for (auto &x : adj[node]) dfs(x, d + 1);
  R[node] = X;
  ++X;
}
int count_smaller(int d, int bound, int arr[]) {
  int sz = dpth[d].size();
  int s = 0, e = sz - 1;
  int res = -1;
  while (s <= e) {
    int mid = (s + e) / 2;
    if (arr[dpth[d][mid]] < bound) {
      s = mid + 1;
      res = mid;
    } else
      e = mid - 1;
  }
  return res;
}
map<pair<pair<int, int>, int>, int> cashe;
int main() {
  scanf("%d", &n);
  adj.resize(n + 1);
  map<string, int> mname;
  for (int i = 1; i <= n; ++i) {
    int parent;
    cin >> name[i] >> parent;
    mname[name[i]] = 0;
    adj[parent].push_back(i);
  }
  int ind = 0;
  for (auto &x : mname) x.second = ind, ++ind;
  for (int i = 1; i <= n; ++i) nameid[i] = mname[name[i]];
  dfs(0, 0);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    int v, k;
    scanf("%d %d", &v, &k);
    int reldpth = ndpth[v] + k;
    if (reldpth >= 100005) {
      printf("%d\n", 0);
      continue;
    }
    int low, high, sz = dpth[reldpth].size();
    low = count_smaller(reldpth, L[v], L) + 1;
    high = count_smaller(reldpth, R[v], R);
    if (low == sz || high == -1) {
      printf("%d\n", 0);
      continue;
    }
    pair<pair<int, int>, int> state = make_pair(make_pair(low, high), reldpth);
    if (cashe.count(state)) {
      printf("%d\n", cashe[state]);
      continue;
    }
    int res = 0;
    for (int j = low; j <= high; ++j) {
      int x = nameid[dpth[reldpth][j]];
      if (!used[x]) {
        used[x] = true;
        ++res;
      }
    }
    for (int j = low; j <= high; ++j) {
      int x = nameid[dpth[reldpth][j]];
      used[x] = false;
    }
    cashe[state] = res;
    printf("%d\n", res);
  }
  return 0;
}
