Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100500;
const int LOGN = 17;
vector<int> sons[N];
int depth[N], jmp[N][LOGN], n, sameAs[N], sdfsdf[N], answer[N];
set<pair<int, int> > start[N], current;
set<int> roots;
void dfs(int v) {
  for (const auto &st : start[v]) {
    auto lb = current.lower_bound(make_pair(st.first, -1));
    if (lb == current.end() or lb->first != st.first) {
      current.insert(st);
    } else {
      answer[st.second] = answer[lb->second];
      sameAs[st.second] = lb->second;
    }
  }
  auto check = current.lower_bound(make_pair(depth[v], -1));
  if (check != current.end() and check->first == depth[v])
    answer[check->second]++;
  for (const int &son : sons[v]) {
    dfs(son);
  }
  for (const auto &st : start[v]) {
    if (sameAs[st.second] > 0) {
      answer[st.second] = answer[sameAs[st.second]] - answer[st.second];
      sameAs[st.second] = 0;
    }
    current.erase(st);
  }
}
void initJump() {
  for (int i = 1; i <= n; i++) {
    jmp[i][0] = sdfsdf[i];
  }
  for (int r = 1; r < LOGN; r++) {
    for (int i = 1; i <= n; i++) jmp[i][r] = jmp[jmp[i][r - 1]][r - 1];
  }
}
int calcDepth(int v) {
  if (sdfsdf[v] == 0) return 0;
  int ret = 0;
  for (int r = LOGN - 1; r >= 0; r--) {
    if (sdfsdf[jmp[v][r]] != 0) {
      ret += 1 << r;
      v = jmp[v][r];
    }
  }
  return ret + 1;
}
int jumpPrev(int v, int d) {
  for (int r = LOGN - 1; r >= 0; r--) {
    if (d >= (1 << r)) {
      v = jmp[v][r];
      d -= (1 << r);
    }
  }
  return v;
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int parent;
    cin >> parent;
    if (parent == 0) {
      roots.insert(i);
      sdfsdf[i] = 0;
    } else {
      sons[parent].push_back(i);
      sdfsdf[i] = parent;
    }
  }
  initJump();
  for (int i = 1; i <= n; i++) {
    depth[i] = calcDepth(i);
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int v, d;
    cin >> v >> d;
    if (depth[v] < d)
      answer[i] = 1;
    else {
      int u = jumpPrev(v, d);
      start[u].insert(make_pair(depth[v], i));
    }
  }
  for (const int &root : roots) {
    current.clear();
    dfs(root);
  }
  for (int i = 1; i <= q; i++) cout << answer[i] - 1 << " ";
  return 0;
}