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 N = 1e5 + 5;
int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, a[N], Q, x, root[N];
char ch[30];
map<string, int> Hash;
int m;
int id[N];
set<int> Set[N];
vector<pair<int, long long> > q[N];
int ans[N];
struct edge {
  int v, ne;
} e[N << 1];
int cnt, h[N];
inline void ins(int u, int v) {
  e[++cnt] = (edge){v, h[u]};
  h[u] = cnt;
}
int size[N], mx[N], deep[N], big[N];
void dfs(int u) {
  size[u] = 1;
  for (int i = h[u]; i; i = e[i].ne) {
    deep[e[i].v] = deep[u] + 1;
    dfs(e[i].v);
    size[u] += size[e[i].v];
    if (size[e[i].v] > size[mx[u]]) mx[u] = e[i].v;
  }
}
pair<int, long long> f[N];
void update(int u, int val) {
  if (val == 1)
    Set[deep[u]].insert(id[u]);
  else
    Set[deep[u]].erase(id[u]);
  for (int i = h[u]; i; i = e[i].ne)
    if (!big[e[i].v]) update(e[i].v, val);
}
inline int cal(int d) { return d > n ? 0 : Set[d].size(); }
void dfs(int u, int keep) {
  for (int i = h[u]; i; i = e[i].ne)
    if (e[i].v != mx[u]) dfs(e[i].v, 0);
  if (mx[u]) dfs(mx[u], 1), big[mx[u]] = 1;
  update(u, 1);
  for (int i = 0; i < (int)q[u].size(); i++)
    ans[q[u][i].first] = cal(deep[u] + q[u][i].second);
  big[mx[u]] = 0;
  if (!keep) update(u, -1);
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    scanf("%s", ch + 1);
    x = read();
    string str(ch + 1);
    if (Hash.count(str))
      id[i] = Hash[str];
    else
      Hash[str] = id[i] = ++m;
    if (x == 0)
      root[++root[0]] = i;
    else
      ins(x, i);
  }
  Q = read();
  for (int i = 1; i <= Q; i++) x = read(), q[x].push_back(make_pair(i, read()));
  for (int i = 1; i <= root[0]; i++) dfs(root[i]), dfs(root[i], 0);
  for (int i = 1; i <= Q; i++) printf("%d\n", ans[i]);
}
