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;
inline string getStr() {
  char ch[1000];
  scanf("%s", ch);
  return ch;
}
inline long long in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
const long long MAX_N = 1e5 + 100;
const long long MAX_LG = 20;
long long root[MAX_N][MAX_LG];
long long level[MAX_N];
vector<vector<long long>> graph(MAX_N);
vector<long long> lv[MAX_N];
long long st[MAX_N], en[MAX_N];
long long cur = 1;
void dfs(long long node, long long pr = 0, long long lvl = 0) {
  root[node][0] = pr;
  st[node] = cur;
  level[node] = lvl;
  lv[lvl].push_back(cur);
  for (long long x = 1; x < MAX_LG; x++)
    root[node][x] = root[root[node][x - 1]][x - 1];
  for (long long i = 0; i < graph[node].size(); i++) {
    cur++;
    dfs(graph[node][i], node, lvl + 1);
  }
  en[node] = cur;
}
long long lca(long long p, long long k) {
  for (long long x = 0; x < MAX_LG && (1 << x) <= k; x++) {
    if (k & (1 << x)) p = root[p][x];
  }
  return p;
}
int32_t main() {
  long long n = in();
  for (long long i = 0; i < n; i++) {
    long long par = in();
    graph[par].push_back(i + 1);
  }
  dfs(0);
  long long q = in();
  while (q--) {
    long long person, pth;
    person = in(), pth = in();
    long long Lc = lca(person, pth);
    if (Lc == 0) {
      cout << "0 ";
      continue;
    }
    long long start = st[Lc];
    long long end = en[Lc];
    long long findLast =
        upper_bound(lv[level[person]].begin(), lv[level[person]].end(), end) -
        lv[level[person]].begin();
    long long findFirst =
        upper_bound(lv[level[person]].begin(), lv[level[person]].end(), start) -
        lv[level[person]].begin();
    printf("%I64d ", findLast - findFirst - 1);
  }
  cout << "\n";
}
