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;
const int maxn = int(1e5);
struct Edge {
  int from, to;
  int next;
} eage[maxn + 5];
int tot = 0;
int head[maxn + 5];
void add(int x, int y) {
  eage[++tot].from = x;
  eage[tot].to = y;
  eage[tot].next = head[x];
  head[x] = tot;
}
int n, m;
int deep[maxn + 5];
bool taken[maxn + 5];
pair<int, int> pos[maxn + 5];
vector<int> a[maxn + 5];
int dfs_clock = 0;
void dfs(int k, int dep) {
  deep[k] = dep;
  taken[k] = true;
  pos[k].first = ++dfs_clock;
  a[dep].push_back(dfs_clock);
  for (int i = head[k]; i; i = eage[i].next) dfs(eage[i].to, dep + 1);
  pos[k].second = ++dfs_clock;
  a[dep].push_back(dfs_clock);
}
int main() {
  scanf("%d", &n);
  for (int x, i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x) add(x, i);
  }
  scanf("%d", &m);
  for (int i = 1; i <= n; i++)
    if (!taken[i]) dfs(i, 0);
  for (int tt = 1; tt <= m; tt++) {
    int v, p;
    scanf("%d%d", &v, &p);
    int ndep = deep[v] - p;
    pair<int, int> cur = pos[v];
    if (ndep < 0) {
      printf("0 ");
      continue;
    }
    vector<int>::iterator t1, f1, t2, f2;
    t1 = upper_bound(a[ndep].begin(), a[ndep].end(), cur.first);
    f1 = t1 - 1;
    f2 = upper_bound(a[deep[v]].begin(), a[deep[v]].end(), *f1);
    t2 = upper_bound(a[deep[v]].begin(), a[deep[v]].end(), *t1) - 1;
    printf("%d ", (t2 - f2 + 1) / 2 - 1);
  }
  return 0;
}
