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 + 10;
int n, m;
int sz[N];
vector<int> e[N];
vector<pair<int, int> > query[N];
int ans[N];
int son[N];
int v[N], k[N];
int color[N], p[N];
map<string, int> mp;
int tot;
int get(string s) {
  if (!mp.count(s)) mp[s] = ++tot;
  return mp[s];
}
void dfs1(int v, int p) {
  sz[v] = 1;
  for (auto u : e[v])
    if (u != p) {
      dfs1(u, v);
      sz[v] += sz[u];
      if (sz[son[v]] < sz[u]) son[v] = u;
    }
}
void mrg(set<int> &a, set<int> &b) {
  if (a.size() < b.size()) swap(a, b);
  for (auto i : b) a.insert(i);
}
void mrg(deque<set<int> > *&a, deque<set<int> > *&b) {
  if (a == 0) a = b;
  for (int i = a->size(); i < b->size(); ++i) (*a).push_back(set<int>());
  for (int i = 0; i < b->size(); ++i) {
    mrg((*a)[i], (*b)[i]);
  }
}
deque<set<int> > *dfs(int v, int p) {
  deque<set<int> > *nw = 0;
  if (son[v]) nw = dfs(son[v], v);
  for (auto u : e[v])
    if (u != p && u != son[v]) {
      auto it = dfs(u, v);
      mrg(nw, it);
    }
  if (!nw) nw = new deque<set<int> >();
  set<int> st;
  st.insert(color[v]);
  nw->push_front(st);
  for (auto t : query[v]) {
    if (t.first >= nw->size())
      ans[t.second] = 0;
    else
      ans[t.second] = (*nw)[t.first].size();
  }
  return nw;
}
int main() {
  scanf("%d", &n);
  vector<int> vv;
  for (int i = 1; i <= n; ++i) {
    char s[30];
    scanf("%s%d", s, p + i);
    color[i] = get(s);
    if (p[i] == 0)
      vv.push_back(i);
    else
      e[p[i]].push_back(i);
  }
  for (auto rt : vv) dfs1(rt, 0);
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", v + i, k + i);
    query[v[i]].push_back(make_pair(k[i], i));
  }
  for (auto rt : vv) dfs(rt, 0);
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}
