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>
const int N = 5e5 + 30;
int read() {
  char ch = getchar();
  int s = 0;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + ch - '0', ch = getchar();
  return s;
}
int n, m, Son, cnt, x, y;
int c[N], siz[N], son[N], ans[N], dep[N], tot[N];
int f[2 * N], nx[2 * N], to[2 * N], F[N][22];
struct question {
  int dep, id;
};
std::vector<question> q[N];
std::string s[N];
std::map<std::string, int> mp[N];
void dfs1(int p, int fa) {
  siz[p] = 1;
  F[p][0] = fa;
  dep[p] = dep[fa] + 1;
  for (int i = 1; i < 20; ++i) F[p][i] = F[F[p][i - 1]][i - 1];
  for (int i = f[p]; i; i = nx[i]) {
    if (to[i] == fa) continue;
    dfs1(to[i], p), siz[p] += siz[to[i]];
    if (siz[to[i]] > siz[son[p]]) son[p] = to[i];
  }
}
void add(int p, int fa, int val) {
  mp[dep[p]][s[p]] += val;
  if (mp[dep[p]][s[p]] == 1 && val == 1) tot[dep[p]]++;
  if (mp[dep[p]][s[p]] == 0) tot[dep[p]]--;
  for (int i = f[p]; i; i = nx[i])
    if (to[i] != Son && to[i] != fa) add(to[i], p, val);
}
void dfs2(int p, int fa, int opt) {
  for (int i = f[p]; i; i = nx[i])
    if (to[i] != son[p] && to[i] != fa) dfs2(to[i], p, 0);
  if (son[p]) dfs2(son[p], p, 1), Son = son[p];
  add(p, fa, 1);
  Son = 0;
  for (int i = 0; i < q[p].size(); ++i)
    ans[q[p][i].id] = tot[q[p][i].dep + dep[p]];
  if (!opt) add(p, fa, -1);
}
int main() {
  n = read() + 1;
  for (int i = 2; i <= n; ++i)
    std::cin >> s[i], x = read() + 1, nx[++nx[0]] = f[x], to[nx[0]] = i,
                      f[x] = nx[0];
  m = read();
  dfs1(1, 0);
  for (int i = 1; i <= m; ++i) {
    x = read() + 1, y = read();
    q[x].push_back((question){y, i});
  }
  dfs2(1, 0, 0);
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}
