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;
long long read() {
  long long a = 0, c = getchar(), s = 0;
  while (!isdigit(c)) s |= c == '-', c = getchar();
  while (isdigit(c)) a = a * 10 + c - 48, c = getchar();
  return s ? -a : a;
}
const int N = 2e5 + 1;
const int BASE1 = 27;
const int BASE2 = 29;
const int MOD1 = 1e9 + 7;
const int MOD2 = 1e9 + 9;
struct Query {
  int id, val;
};
int n, m;
char s[21];
long long name[N];
int len;
long long val[N];
int id[N];
int top, fi[N], ne[N << 1], to[N << 1];
int fa[N], deep[N], size[N], maxson[N];
unordered_map<int, bool> have[N];
vector<Query> query[N];
int res[N];
long long GetHash(char *s);
void Add(int u, int v);
void GetInfo(int t);
void GetRes(int t);
void GetTree(int rt, int t, int v);
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s);
    name[i] = GetHash(s);
    val[++len] = name[i];
    Add(read(), i);
  }
  sort(val + 1, val + len + 1);
  len = unique(val + 1, val + len + 1) - val - 1;
  for (int i = 1; i <= n; ++i)
    id[i] = upper_bound(val + 1, val + len + 1, name[i]) - val - 1;
  m = read();
  for (int i = 1; i <= m; ++i) {
    int v = read(), k = read();
    query[v].push_back((Query){i, k});
  }
  GetInfo(0);
  GetRes(0);
  for (int i = 1; i <= m; ++i) printf("%d\n", res[i]);
  return 0;
}
long long GetHash(char *s) {
  long long v1 = 0, v2 = 0;
  for (int i = 0; s[i]; ++i) {
    ((v1 += s[i] - 'a' + 1) *= BASE1) %= MOD1;
    ((v2 += s[i] - 'a' + 1) *= BASE2) %= MOD2;
  }
  return v1 * MOD2 + v2;
}
void Add(int u, int v) { ne[++top] = fi[u], fi[u] = top, to[top] = v; }
void GetInfo(int t) {
  size[t] = 1;
  for (int i = fi[t]; i; i = ne[i]) {
    if (to[i] == fa[t]) continue;
    fa[to[i]] = t;
    deep[to[i]] = deep[t] + 1;
    GetInfo(to[i]);
    size[t] += size[to[i]];
    if (size[to[i]] > size[maxson[t]]) maxson[t] = to[i];
  }
}
void GetRes(int t) {
  for (int i = fi[t]; i; i = ne[i]) {
    if (to[i] == fa[t] || to[i] == maxson[t]) continue;
    GetRes(to[i]);
    GetTree(t, to[i], -1);
  }
  if (maxson[t]) GetRes(maxson[t]);
  GetTree(t, t, 1);
  for (size_t i = 0; i < query[t].size(); ++i) {
    Query tq = query[t].at(i);
    res[tq.id] = have[deep[t] + tq.val].size();
  }
}
void GetTree(int rt, int t, int v) {
  if (v == -1)
    have[deep[t]].erase(id[t]);
  else
    have[deep[t]][id[t]] = true;
  for (int i = fi[t]; i; i = ne[i]) {
    if (to[i] == fa[t] || to[i] == maxson[rt]) continue;
    GetTree(rt, to[i], v);
  }
}
