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 + 20;
struct Edge {
  int ver, next;
} e[N << 1];
int n, m, t, cnt, Head[N], sz[N], son[N], dep[N], c[N], tot[N], Ans[N];
char str[N];
map<string, int> id;
multiset<int> s[N];
vector<pair<int, int> > q[N];
vector<int> rt;
string name[N];
inline void Insert(int x, int y) { e[++t] = (Edge){y, Head[x]}, Head[x] = t; }
inline void Dfs(int x) {
  sz[x] = 1, son[x] = 0;
  for (int i = Head[x], y; i; i = e[i].next)
    dep[y = e[i].ver] = dep[x] + 1, Dfs(y), sz[x] += sz[y],
            (sz[y] <= sz[son[x]] ?: son[x] = y);
}
inline void Add(int x) {
  if (s[dep[x]].count(c[x]) == 0) tot[dep[x]]++;
  s[dep[x]].insert(c[x]);
}
inline void Del(int x) {
  if (s[dep[x]].count(c[x]) == 1) tot[dep[x]]--;
  s[dep[x]].erase(s[dep[x]].find(c[x]));
}
inline void Modify(int x, int p, int op) {
  if (op == 1) Add(x);
  if (op == -1) Del(x);
  for (int i = Head[x], y; i; i = e[i].next)
    if ((y = e[i].ver) != p) Modify(y, p, op);
}
inline void Dsu(int x, int op) {
  for (int i = Head[x], y; i; i = e[i].next)
    if ((y = e[i].ver) != son[x]) Dsu(y, 0);
  if (son[x]) Dsu(son[x], 1);
  Modify(x, son[x], 1);
  for (int i = 0; i < q[x].size(); i++)
    if (dep[x] + q[x][i].first <= n)
      Ans[q[x][i].second] = tot[dep[x] + q[x][i].first];
  if (op == 0) Modify(x, 0, -1);
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1, fa, len; i <= n; i++) {
    scanf("%s", str + 1), len = strlen(str + 1);
    for (int j = 1; j <= len; j++) name[i] += str[j];
    scanf("%d", &fa), id[name[i]] = 1;
    if (fa)
      Insert(fa, i);
    else
      rt.push_back(i);
  }
  map<string, int>::iterator it = id.begin();
  for (; it != id.end(); it++) (it->second) = ++cnt;
  for (int i = 1; i <= n; i++) c[i] = id[name[i]];
  scanf("%d", &m);
  for (int i = 1, v, k; i <= m; i++)
    scanf("%d%d", &v, &k), q[v].push_back(pair<int, int>(k, i));
  for (int i = 0; i < rt.size(); i++) dep[rt[i]] = 1, Dfs(rt[i]), Dsu(rt[i], 0);
  for (int i = 1; i <= m; i++) printf("%d\n", Ans[i]);
  return 0;
}
