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;
int fa[18][100005], dep[100005], n, q, t, ia[100005], ib[100005];
vector<int> ve[100005], vec[100005], aq;
void update() {
  for (int i = 0; i <= 17; i++) {
    for (int j = 1; j <= n; j++) {
      if (fa[i][j] != -1) fa[i + 1][j] = fa[i][fa[i][j]];
    }
  }
}
void dfs(int pt, int f) {
  if (f == -1)
    dep[pt] = 0;
  else
    dep[pt] = dep[f] + 1;
  ia[pt] = t++;
  vec[dep[pt]].push_back(ia[pt]);
  for (int i = 0; i < ve[pt].size(); i++) dfs(ve[pt][i], pt);
  ib[pt] = t;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  memset(fa, -1, sizeof(fa));
  for (int i = 1; i <= n; i++) {
    cin >> fa[0][i];
    if (fa[0][i] == 0) {
      fa[0][i] = -1;
      continue;
    } else
      ve[fa[0][i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (fa[0][i] == -1) {
      dfs(i, -1);
    }
  }
  update();
  cin >> q;
  bool sp = false;
  while (q--) {
    int v, p;
    cin >> v >> p;
    if (dep[v] < p) {
      aq.push_back(0);
      continue;
    }
    int va = v, de = dep[v];
    for (int i = 17; i >= 0; i--)
      if (p & (1 << i)) va = fa[i][va];
    int ans = int(lower_bound(vec[de].begin(), vec[de].end(), ib[va]) -
                  vec[de].begin()) -
              int(lower_bound(vec[de].begin(), vec[de].end(), ia[va]) -
                  vec[de].begin()) -
              1;
    aq.push_back(ans);
  }
  for (int i = 0; i < aq.size(); i++) {
    if (i > 0) cout << " ";
    cout << aq[i];
  }
  cout << endl;
  return 0;
}