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 MAX = (1e5) + 10;
int lf[MAX], rg[MAX], lev[MAX], bigChild[MAX], sz[MAX], dfa[MAX], ti;
int col[MAX];
vector<int> mat[MAX];
vector<pair<int, int> > query[MAX];
set<int> sota[MAX];
void dfs(int p, int pa) {
  lf[p] = ++ti;
  sz[p] = 1;
  dfa[ti] = p;
  int mx = -1;
  bigChild[p] = -1;
  for (auto x : mat[p]) {
    if (x != pa) {
      lev[x] = lev[p] + 1;
      dfs(x, p);
      sz[p] += sz[x];
      if (sz[x] > mx) mx = sz[x], bigChild[p] = x;
    }
  }
  rg[p] = ti;
}
void operationOnNode(int p, bool shouldDelete) {
  for (int i = lf[p]; i <= rg[p]; i++) {
    int v = dfa[i];
    if (shouldDelete)
      sota[lev[v]].erase(col[v]);
    else
      sota[lev[v]].insert(col[v]);
  }
}
int dekh[MAX];
void dsu(int p, int pa, int keep) {
  for (auto x : mat[p])
    if (x != pa && x != bigChild[p]) dsu(x, p, 0);
  if (bigChild[p] != -1) dsu(bigChild[p], p, 1);
  sota[lev[p]].insert(col[p]);
  for (int x : mat[p]) {
    if (x == bigChild[p]) continue;
    operationOnNode(x, 0);
  }
  for (pair<int, int> x : query[p]) {
    int i = x.second, l = x.first;
    if (lev[p] + l < MAX) dekh[i] = sota[(lev[p] + l)].size();
  }
  if (!keep) operationOnNode(p, 1);
}
map<unsigned long long, int> giveId;
int giveIdForULL(unsigned long long v) {
  if (giveId[v] == 0) giveId[v] = giveId.size();
  return giveId[v];
}
unsigned long long getStringId(string str) {
  int n = str.size();
  unsigned long long hashValue = 0, ba = 1;
  for (int i = 0; i < n; i++) {
    int x = (str[i] - 'a') + 1;
    hashValue += (x * ba);
    ba *= 29;
  }
  return giveIdForULL(hashValue);
}
int main() {
  ios::sync_with_stdio(false);
  int n, q, x, y, root;
  string str;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> str >> y;
    col[i] = getStringId(str);
    mat[y].push_back(i);
  }
  root = 0;
  cin >> q;
  for (int i = 0; i <= q - 1; i++) {
    cin >> x >> y;
    query[x].push_back({y, i});
  }
  dfs(root, root);
  dsu(root, root, 0);
  for (int i = 0; i <= q - 1; i++) cout << dekh[i] << endl;
  return 0;
}
