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;
struct node {
  int lab, lvl, pos;
  node(int a = 0, int b = 0, int c = 0) { lab = a, lvl = b, pos = c; }
  bool operator<(const node &ot) const {
    if (lvl == ot.lvl) return pos < ot.pos;
    return lvl < ot.lvl;
  }
} nds[100005];
int n;
vector<int> ady[100005];
bitset<100005> mk;
int lv[100005], s[100005], first[100005], id;
void dfs(int u) {
  nds[u] = node(u, lv[u], ++id);
  first[u] = id;
  mk[u] = 1;
  s[u] = 1;
  for (auto v : ady[u]) {
    if (!mk[v]) {
      lv[v] = lv[u] + 1;
      dfs(v);
      s[u] += s[v];
    }
  }
  mk[u] = 0;
}
int bs2(int k, int v) {
  int lvl = lv[v];
  int l, r;
  l = 1, r = n;
  while (l != r) {
    int m = (l + r) >> 1;
    if (nds[m].lvl < lvl)
      l = m + 1;
    else
      r = m;
  }
  int L = l;
  l = 1, r = n;
  while (l != r) {
    int m = (l + r) >> 1;
    if (nds[m].lvl <= lvl)
      l = m + 1;
    else
      r = m;
  }
  int R = (nds[r].lvl > lvl ? r - 1 : r);
  l = L, r = R;
  while (l != r) {
    int m = (l + r) >> 1;
    if (first[k] > nds[m].pos)
      l = m + 1;
    else
      r = m;
  }
  int fir = l;
  l = L, r = R;
  while (l != r) {
    int m = (l + r) >> 1;
    if (first[k] + s[k] - 1 >= nds[m].pos)
      l = m + 1;
    else
      r = m;
  }
  int sec = (first[k] + s[k] - 1 < nds[l].pos || l > R ? l - 1 : l);
  return sec - fir + 1;
}
int bs(int v, int lvl) {
  int l, r;
  l = 1, r = n;
  while (l != r) {
    int m = (l + r) >> 1;
    if (nds[m].lvl < lvl)
      l = m + 1;
    else
      r = m;
  }
  int L = l;
  l = L, r = n;
  while (l != r) {
    int m = (l + r) >> 1;
    if (nds[m].lvl <= lvl)
      l = m + 1;
    else
      r = m;
  }
  int R = (nds[l].lvl > lvl ? l - 1 : l);
  l = L, r = R;
  while (l != r) {
    int m = (l + r) >> 1;
    if (nds[m].pos < first[v])
      l = m + 1;
    else
      r = m;
  }
  int k = (nds[l].pos > first[v] ? nds[l - 1].lab : nds[l].lab);
  return bs2(k, v);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<int> roots;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p == 0)
      roots.push_back(i);
    else
      ady[p].push_back(i);
  }
  for (int u : roots) dfs(u);
  sort(nds + 1, nds + n + 1);
  int q;
  cin >> q;
  while (q--) {
    int v, p;
    cin >> v >> p;
    if (lv[v] - p < 0)
      cout << 0 << " ";
    else
      cout << bs(v, lv[v] - p) - 1 << " ";
  }
  return 0;
}