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;
vector<int> graph[100005];
vector<pair<int, int> > query[100005], one[100005];
int level[100005], visited[100005], arr[100005], ans[100005], cnt[100005],
    temp[100005];
void dfs1(int v, int l) {
  visited[v] = 1;
  level[v] = l;
  arr[l] = v;
  for (int i = 0; i < graph[v].size(); i++) {
    int u = graph[v][i];
    if (!visited[u]) {
      dfs1(u, l + 1);
    }
  }
  for (int j = 0; j < query[v].size(); j++) {
    int u = query[v][j].first, index = query[v][j].second;
    if (u > l) continue;
    int ind = arr[l - u];
    one[ind].push_back(make_pair(level[v], index));
  }
}
void dfs2(int v) {
  visited[v] = 2;
  cnt[level[v]]++;
  for (int i = 0; i < one[v].size(); i++) {
    int u = one[v][i].first, d = one[v][i].second;
    temp[d] = cnt[u];
  }
  for (int i = 0; i < graph[v].size(); i++) {
    int u = graph[v][i];
    if (visited[u] == 1) {
      dfs2(u);
    }
  }
  for (int i = 0; i < one[v].size(); i++) {
    int u = one[v][i].first, d = one[v][i].second;
    ans[d] = cnt[u] - temp[d] - 1;
  }
}
int main() {
  int n, a, m, v, p;
  scanf("%d", &n);
  vector<int> start;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    if (a == 0) {
      start.push_back(i);
      continue;
    }
    graph[a].push_back(i);
    graph[i].push_back(a);
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &v, &p);
    query[v].push_back(make_pair(p, i));
  }
  for (int i = 0; i < start.size(); i++) {
    dfs1(start[i], 0);
    dfs2(start[i]);
  }
  for (int i = 0; i < m; i++) printf("%d ", ans[i]);
  return 0;
}
