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> mat[500010];
vector<pair<int, int> > quer[500010];
int lf[500010], rg[500010], lev[500010], bigC[500010], sz[500010], dfa[500010],
    ti;
void dfs(int p, int pa) {
  lf[p] = ++ti;
  sz[p] = 1;
  dfa[ti] = p;
  int mx = -1;
  bigC[p] = -1;
  for (auto x : mat[p]) {
    if (x == pa) continue;
    lev[x] = lev[p] + 1;
    dfs(x, p);
    sz[p] += sz[x];
    if (sz[x] > mx) mx = sz[x], bigC[p] = x;
  }
  rg[p] = ti;
}
int prin[500010], col[500010];
set<int> occ[500010];
void update(int p, int v) {
  int l = lev[p];
  if (v == 1)
    occ[l].insert(col[p]);
  else
    occ[l].erase(col[p]);
  return;
}
int dsu(int p, int pa, int keep) {
  for (auto x : mat[p])
    if (x != pa && x != bigC[p]) dsu(x, p, 0);
  if (bigC[p] != -1) dsu(bigC[p], p, 1);
  update(p, 1);
  for (auto x : mat[p]) {
    if (x == pa || x == bigC[p]) continue;
    for (int i = lf[x]; i <= rg[x]; i++) update(dfa[i], 1);
  }
  for (auto x : quer[p]) {
    int l = lev[p] + x.first;
    if (l < 500010) prin[x.second] = occ[l].size();
  }
  if (!keep)
    for (int i = lf[p]; i <= rg[p]; i++) update(dfa[i], -1);
}
unsigned int getHash(string str) {
  int n = str.size();
  unsigned int ret = 0, ba = 31, c;
  for (int i = 0; i <= n - 1; i++) {
    c = str[i] - 'a';
    c++;
    ret += (ba * c);
    ba *= 31;
  }
  return ret;
}
map<unsigned int, int> mopa;
string str;
int getId(string str) {
  unsigned int v = getHash(str);
  if (mopa[v] == 0) mopa[v] = mopa.size();
  return mopa[v];
}
vector<int> Root;
int main() {
  ios::sync_with_stdio(false);
  int n, u, v, q, root;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    u = i;
    cin >> str >> v;
    if (v) mat[u].push_back(v);
    if (v) mat[v].push_back(u);
    col[i] = getId(str);
    if (v == 0) Root.push_back(u);
  }
  cin >> q;
  for (int i = 0; i <= q - 1; i++) {
    cin >> u >> v;
    quer[u].push_back({v, i});
  }
  for (auto root : Root) {
    dfs(root, root);
    dsu(root, root, 0);
  }
  for (int i = 0; i <= q - 1; i++) cout << prin[i] << endl;
  return 0;
}
