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;
template <class T, class L>
bool smax(T& x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
bool smin(T& x, L y) {
  return x > y ? (x = y, 1) : 0;
}
const int maxn = 2e5 + 17;
unordered_map<string, int> gooni[maxn];
int ans[maxn], n, sz[maxn], h[maxn], q;
string s[maxn];
bool mark[maxn];
vector<int> g[maxn];
vector<pair<int, int> > as[maxn];
void OstadLovesInit(int v = 0) {
  sz[v] = 1;
  for (auto u : g[v]) h[u] = h[v] + 1, OstadLovesInit(u), sz[v] += sz[u];
}
void rem(int v) {
  auto it = gooni[h[v]].find(s[v]);
  it->second--;
  if (!it->second) gooni[h[v]].erase(it);
  for (auto u : g[v])
    if (!mark[u]) rem(u);
}
void add(int v) {
  gooni[h[v]][s[v]]++;
  for (auto u : g[v])
    if (!mark[u]) add(u);
}
void dfs(int v = 0, bool cl = 0) {
  int bigi, big = 0;
  for (auto u : g[v])
    if (smax(big, sz[u])) bigi = u;
  for (auto u : g[v])
    if (u != bigi) dfs(u, 1);
  if (big) dfs(bigi), mark[bigi] = 1;
  add(v);
  for (auto q : as[v]) ans[q.second] = gooni[h[v] + q.first].size();
  if (big) mark[bigi] = 0;
  if (cl) rem(v);
}
int main() {
  int u, v;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    scanf("%d", &u);
    g[u].push_back(i);
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d", &u);
    scanf("%d", &v);
    as[u].push_back(make_pair(v, i));
  }
  OstadLovesInit();
  dfs();
  for (int i = 0; i < q; i++) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}
