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 visited[100005], level[100005], parent[100005], ancestor[20][100005],
    result[100005];
vector<int> graph[100005], root;
map<int, int>* M[100005];
map<int, int>::iterator it;
vector<pair<int, int> > query[100005];
void dfs(int node) {
  int i, temp;
  for (i = 0; i < graph[node].size(); i++) {
    temp = graph[node][i];
    if (visited[temp] == 0) {
      visited[temp] = 1;
      level[temp] = level[node] + 1;
      parent[temp] = node;
      dfs(temp);
    }
  }
}
void dfs2(int node, int parent) {
  int i, temp, maxi = -1, idx = -1;
  for (i = 0; i < graph[node].size(); i++) {
    temp = graph[node][i];
    if (visited[temp] == 0) {
      visited[temp] = 1;
      dfs2(temp, node);
      if ((int)M[temp]->size() > maxi) {
        maxi = (int)M[temp]->size();
        idx = temp;
      }
    }
  }
  if (idx != -1) M[node] = M[idx];
  (*M[node])[level[node]]++;
  for (i = 0; i < graph[node].size(); i++) {
    temp = graph[node][i];
    if (temp != idx && temp != parent) {
      for (it = M[temp]->begin(); it != M[temp]->end(); ++it)
        (*M[node])[it->first] += it->second;
      free(M[temp]);
    }
  }
  for (i = 0; i < query[node].size(); i++) {
    pair<int, int> bar = query[node][i];
    result[bar.second] = (*M[node])[bar.first];
  }
  return;
}
void preCompute(int n) {
  int i, j;
  for (i = 1; i <= n; i++) ancestor[0][i] = parent[i];
  for (i = 1; i < 20; i++)
    for (j = 1; j <= n; j++)
      ancestor[i][j] = ancestor[i - 1][ancestor[i - 1][j]];
  return;
}
int getAncestor(int node, int depth) {
  int i, ret = node;
  if (depth == 0) return ret;
  for (i = 0; i < 20; i++)
    if (((depth >> i) & 1)) ret = ancestor[i][ret];
  return ret;
}
int main() {
  int i, j, t, n, m, k, l, r, temp, mini, maxi, flag, cnt;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    M[i] = new map<int, int>;
    scanf("%d", &k);
    if (k != 0) {
      graph[i].push_back(k);
      graph[k].push_back(i);
    } else
      root.push_back(i);
  }
  for (i = 0; i < root.size(); i++) {
    temp = root[i];
    visited[temp] = 1;
    level[temp] = 0;
    dfs(temp);
  }
  preCompute(n);
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &l, &r);
    int anc = getAncestor(l, r);
    if (anc == 0)
      result[i] = 1;
    else
      query[anc].push_back(make_pair(level[anc] + r, i));
  }
  for (i = 1; i <= n; i++) visited[i] = 0;
  for (i = 0; i < root.size(); i++) {
    temp = root[i];
    visited[temp] = 1;
    dfs2(temp, 0);
  }
  for (i = 1; i <= m; i++) printf("%d ", result[i] - 1);
  printf("\n");
  return 0;
}