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;
int n;
vector<int> graph[100005];
vector<int> nodesAtLevel[100005];
int rootVertex[100005];
int parent[100005];
int level[100005];
int ancestor[18][100005];
int dfsNumber;
int dfsBegin[100005];
int dfsEnd[100005];
void dfs(int u, int distance, int root) {
  dfsBegin[u] = ++dfsNumber;
  rootVertex[u] = root;
  level[u] = distance;
  nodesAtLevel[distance].push_back(u);
  for (auto v : graph[u]) {
    dfs(v, distance + 1, root);
  }
  dfsEnd[u] = dfsNumber;
}
void lcaPreprocess() {
  int i, j;
  for (i = 1; i <= n; i++) {
    ancestor[0][i] = parent[i];
  }
  for (j = 1; j <= 17; j++) {
    for (i = 1; i <= n; i++) {
      if (ancestor[j - 1] != 0) {
        ancestor[j][i] = ancestor[j - 1][ancestor[j - 1][i]];
      }
    }
  }
}
int getKthAncestor(int u, int k) {
  for (int j = 17; j >= 0; j--) {
    if (level[u] >= (1 << j) && (1 << j) <= k) {
      u = ancestor[j][u];
      k -= (1 << j);
    }
  }
  return (k == 0 ? u : -1);
}
int main() {
  int q, i, j, u, v, p, first, last, mid, firstIndex, lastIndex;
  scanf("%d", &n);
  for (v = 1; v <= n; v++) {
    scanf("%d", &u);
    parent[v] = u;
    if (u != 0) {
      graph[u].push_back(v);
    }
  }
  for (u = 1; u <= n; u++) {
    if (parent[u] == 0) {
      dfs(u, 0, u);
    }
  }
  lcaPreprocess();
  scanf("%d ", &q);
  while (q--) {
    scanf("%d %d", &v, &p);
    int ancestor = getKthAncestor(v, p);
    if (ancestor == -1) {
      cout << "0 ";
    } else {
      firstIndex = 0;
      lastIndex = 0;
      first = 0;
      last = nodesAtLevel[level[ancestor] + p].size() - 1;
      while (first <= last) {
        mid = (first + last) / 2;
        if (dfsBegin[ancestor] <=
                dfsBegin[nodesAtLevel[level[ancestor] + p][mid]] &&
            dfsBegin[nodesAtLevel[level[ancestor] + p][mid]] <=
                dfsEnd[ancestor]) {
          firstIndex = mid;
          last = mid - 1;
        } else if (dfsBegin[nodesAtLevel[level[ancestor] + p][mid]] >
                   dfsEnd[ancestor]) {
          last = mid - 1;
        } else {
          first = mid + 1;
        }
      }
      first = 0;
      last = nodesAtLevel[level[ancestor] + p].size() - 1;
      while (first <= last) {
        mid = (first + last) / 2;
        if (dfsBegin[ancestor] <=
                dfsBegin[nodesAtLevel[level[ancestor] + p][mid]] &&
            dfsBegin[nodesAtLevel[level[ancestor] + p][mid]] <=
                dfsEnd[ancestor]) {
          lastIndex = mid;
          first = mid + 1;
        } else if (dfsBegin[nodesAtLevel[level[ancestor] + p][mid]] >
                   dfsEnd[ancestor]) {
          last = mid - 1;
        } else {
          first = mid + 1;
        }
      }
      printf("%d ", (lastIndex - firstIndex + 1) - 1);
    }
  }
  return 0;
}
