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> v[100004];
vector<int> v2[100004];
bool seen[100004];
int par[100004], L[100004], root[100004], bap[100004][17], SZ[100006];
int r, cnt = 1, start = 0;
stack<int> s;
bool seen2[100004];
int ST[100005];
void dfs(int x, int level) {
  if (seen[x]) return;
  seen[x] = 1;
  SZ[x] = 1;
  v2[level].push_back(start);
  ST[x] = start;
  start++;
  L[x] = level;
  for (int i = 0; i < v[x].size(); i++) dfs(v[x][i], level + 1);
  SZ[par[x]] += SZ[x];
}
void counting(int n) {
  for (int i = 1; i <= n; i++) bap[i][0] = par[i];
  for (int i = 1; i <= 16; i++)
    for (int j = 1; j <= n; j++) {
      bap[j][i] = bap[bap[j][i - 1]][i - 1];
    }
}
int go(int x, int diff) {
  int log, ans = 0;
  while (diff != 0) {
    log = log2(diff);
    x = bap[x][log];
    diff -= (1 << log);
  }
  return x;
}
int main() {
  int n, x, q, p;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    par[i] = x;
    v[x].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    if (seen[i] == 0 && par[i] == 0) dfs(i, 0);
  counting(n);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &x, &p);
    int ans = go(x, p);
    if (ans == 0)
      printf("0 ");
    else {
      int r1 = ST[ans], r2 = ST[ans] + SZ[ans] - 1;
      int low =
          lower_bound(v2[L[x]].begin(), v2[L[x]].end(), r1) - v2[L[x]].begin();
      int up =
          upper_bound(v2[L[x]].begin(), v2[L[x]].end(), r2) - v2[L[x]].begin();
      ans = up - low;
      printf("%d ", ans - 1);
    }
  }
  return 0;
}
