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 = 100010;
const int MAXL = 2000010;
int _w;
int n, m, a[MAXN], ans[MAXN];
char s[30];
vector<int> rt, qd[MAXN], qid[MAXN];
namespace Trie {
int ch[MAXL][26], nid;
int ins(char *s) {
  int u = 0;
  for (; *s; ++s) {
    int c = *s - 'a';
    if (!ch[u][c]) ch[u][c] = ++nid;
    u = ch[u][c];
  }
  return u;
}
}  // namespace Trie
namespace G {
int head[MAXN], nxt[MAXN], to[MAXN], eid;
void init() {
  eid = 0;
  memset(head, -1, sizeof head);
}
void adde(int u, int v) { to[eid] = v, nxt[eid] = head[u], head[u] = eid++; }
}  // namespace G
int dep[MAXN], sz[MAXN], son[MAXN];
void prelude(int u, int d) {
  using namespace G;
  dep[u] = d, sz[u] = 1, son[u] = 0;
  for (int i = head[u]; ~i; i = nxt[i]) {
    int v = to[i];
    prelude(v, d + 1);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}
set<int> name[MAXN];
void ins_tree(int u) {
  using namespace G;
  name[dep[u]].insert(a[u]);
  for (int i = head[u]; ~i; i = nxt[i]) ins_tree(to[i]);
}
void clear(int u) {
  using namespace G;
  name[dep[u]].clear();
  for (int i = head[u]; ~i; i = nxt[i]) clear(to[i]);
}
void solve(int u, bool clr) {
  using namespace G;
  for (int i = head[u]; ~i; i = nxt[i])
    if (to[i] != son[u]) solve(to[i], 1);
  if (son[u]) solve(son[u], 0);
  for (int i = head[u]; ~i; i = nxt[i])
    if (to[i] != son[u]) ins_tree(to[i]);
  name[dep[u]].insert(a[u]);
  int q = (int)qd[u].size();
  for (int i = 0; i < q; ++i) {
    int d = qd[u][i], id = qid[u][i];
    if (d <= n) ans[id] = (int)name[d].size();
  }
  if (clr) clear(u);
}
int main() {
  _w = scanf("%d", &n);
  G::init();
  for (int i = 1; i <= n; ++i) {
    int pa;
    _w = scanf("%s%d", s, &pa);
    a[i] = Trie::ins(s);
    if (pa)
      G::adde(pa, i);
    else
      rt.push_back(i);
  }
  for (vector<int>::iterator it = rt.begin(); it != rt.end(); ++it)
    prelude(*it, 1);
  _w = scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int u, d;
    _w = scanf("%d%d", &u, &d);
    d += dep[u];
    qd[u].push_back(d);
    qid[u].push_back(i);
  }
  for (vector<int>::iterator it = rt.begin(); it != rt.end(); ++it)
    solve(*it, 1);
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}
