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 N = 1e5 + 5;
vector<int> Edge[N], Root;
vector<pair<int, int> > Ask[N];
namespace HASH {
int tot;
map<string, int> Hash;
inline int get(string st) {
  if (!Hash.count(st)) Hash[st] = ++tot;
  return Hash[st];
}
}  // namespace HASH
using namespace HASH;
int n, m, c[N];
int Size[N], dep[N], son[N], L[N], R[N], id[N], sum;
inline void dfsA(int x) {
  Size[x] = 1;
  L[x] = ++sum;
  id[sum] = x;
  for (int i = 0; i < Edge[x].size(); i++) {
    int y = Edge[x][i];
    dep[y] = dep[x] + 1;
    dfsA(y);
    Size[x] += Size[y];
    if (Size[y] > Size[son[x]]) son[x] = y;
  }
  R[x] = sum;
}
int ans[N], res[N * 2];
map<int, int> cnt[N * 2];
inline void dfsB(int x, bool keep) {
  for (int i = 0; i < Edge[x].size(); i++) {
    int y = Edge[x][i];
    if (y != son[x]) dfsB(y, 0);
  }
  if (son[x]) dfsB(son[x], 1);
  cnt[dep[x]][c[x]]++;
  if (cnt[dep[x]][c[x]] == 1) res[dep[x]]++;
  for (int i = 0; i < Edge[x].size(); i++) {
    int y = Edge[x][i];
    if (y == son[x]) continue;
    for (int j = L[y]; j <= R[y]; j++) {
      int z = id[j];
      cnt[dep[z]][c[z]]++;
      if (cnt[dep[z]][c[z]] == 1) res[dep[z]]++;
    }
  }
  for (int i = 0; i < Ask[x].size(); i++) {
    int y = Ask[x][i].first, z = Ask[x][i].second;
    ans[z] = res[y];
  }
  if (!keep) {
    for (int i = L[x]; i <= R[x]; i++) {
      int y = id[i];
      cnt[dep[y]][c[y]]--;
      if (!cnt[dep[y]][c[y]]) res[dep[y]]--;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int x, y = 1; y <= n; y++) {
    string name;
    cin >> name >> x;
    c[y] = get(name);
    if (!x)
      Root.push_back(y);
    else
      Edge[x].push_back(y);
  }
  for (int i = 0; i < Root.size(); i++) dfsA(Root[i]);
  cin >> m;
  for (int i = 1, x, y; i <= m; i++) {
    cin >> x >> y;
    Ask[x].push_back(make_pair(dep[x] + y, i));
  }
  for (int i = 0; i < Root.size(); i++) dfsB(Root[i], 0);
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
  return 0;
}
