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 a[17][110000];
vector<int> ch[110000];
vector<int> d[110000];
pair<int, int> posi[110000];
vector<pair<int, int> > q;
int solve(int x, int p) {
  int i, now;
  now = x;
  for (i = 16; i >= 0; i--)
    if (((1 << i) & p) != 0) {
      if (now != -1) now = a[i][now];
    }
  return now;
}
int get(int x, int p) {
  int left, right, mid, tmp, x1, x2;
  pair<int, int> xx;
  xx = posi[x];
  left = -1;
  right = d[xx.first + p].size();
  while (right - left > 1) {
    mid = (left + right) / 2;
    tmp = solve(d[xx.first + p][mid], p);
    if (posi[tmp].second < xx.second)
      left = mid;
    else
      right = mid;
  }
  x1 = right;
  left = -1;
  right = d[xx.first + p].size();
  while (right - left > 1) {
    mid = (left + right) / 2;
    tmp = solve(d[xx.first + p][mid], p);
    if (posi[tmp].second <= xx.second)
      left = mid;
    else
      right = mid;
  }
  x2 = left;
  return x2 - x1 + 1;
}
int main() {
  int i, j, n, x, y, all, now, m;
  scanf("%d", &n);
  for (i = 0; i < n; i++) ch[i].clear();
  for (i = 0; i < n; i++) {
    scanf("%d", &x);
    x--;
    a[0][i] = x;
    if (x != -1) ch[x].push_back(i);
  }
  for (i = 1; i < 17; i++) {
    for (j = 0; j < n; j++) {
      if (a[i - 1][j] == -1)
        a[i][j] = -1;
      else
        a[i][j] = a[i - 1][a[i - 1][j]];
    }
  }
  for (i = 0; i < n; i++) d[i].clear();
  for (i = 0; i < n; i++)
    if (a[0][i] == -1) {
      d[0].push_back(i);
      posi[i] = make_pair(0, d[0].size() - 1);
      q.clear();
      q.push_back(make_pair(0, i));
      while (q.size() > 0) {
        pair<int, int> p = q[q.size() - 1];
        q.pop_back();
        for (j = 0; j < ch[p.second].size(); j++) {
          q.push_back(make_pair(p.first + 1, ch[p.second][j]));
        }
        if (ch[p.second].size() > 0) {
          for (j = ch[p.second].size() - 1; j >= 0; j--) {
            d[p.first + 1].push_back(ch[p.second][j]);
            posi[ch[p.second][j]] =
                make_pair(p.first + 1, d[p.first + 1].size() - 1);
          }
        }
      }
    }
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    if (i != 0) printf(" ");
    scanf("%d%d", &x, &y);
    x--;
    now = solve(x, y);
    if (now == -1) {
      printf("0");
    } else {
      all = get(now, y);
      all--;
      printf("%d", all);
    }
  }
  printf("\n");
}