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 = 100005;
unordered_map<string, int> mp[N];
string str[N];
int depth[N], sz[N], heav[N], co[N], ans[N], dm;
vector<int> adj[N];
vector<bool> big(N);
vector<pair<int, int> > q[N];
void getsz(int s) {
  sz[s] = 1;
  int ma = -1;
  for (auto it : adj[s]) {
    depth[it] = depth[s] + 1;
    dm = max(dm, depth[it]);
    getsz(it);
    sz[s] += sz[it];
    if (sz[it] > ma) {
      ma = sz[it];
      heav[s] = it;
    }
  }
}
void rem(int s) {
  int k = --mp[depth[s]][str[s]];
  if (k == 0) co[depth[s]]--;
  for (auto it : adj[s]) {
    if (!big[it]) rem(it);
  }
}
void add(int s) {
  int k;
  k = ++mp[depth[s]][str[s]];
  if (k == 1) co[depth[s]]++;
  for (auto it : adj[s]) {
    if (!big[it]) {
      add(it);
    }
  }
}
void dfs(int s, bool isbig) {
  for (auto it : adj[s]) {
    if (it != heav[s]) dfs(it, 0);
  }
  if (heav[s] != -1) {
    dfs(heav[s], 1);
    big[heav[s]] = 1;
  }
  add(s);
  if (heav[s] != -1) big[heav[s]] = 0;
  for (int i = 0; i < q[s].size(); i++) {
    if (depth[s] + q[s][i].first <= dm)
      ans[q[s][i].second] = co[depth[s] + q[s][i].first];
    else
      ans[q[s][i].second] = 0;
  }
  if (!isbig) rem(s);
}
int main() {
  int n, r, i, m;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  memset(heav, -1, sizeof(heav));
  for (i = 1; i <= n; i++) {
    cin >> str[i];
    cin >> r;
    adj[r].push_back(i);
  }
  cin >> m;
  for (i = 0; i < m; i++) {
    int v, k;
    cin >> v >> k;
    q[v].push_back({k, i});
  }
  getsz(0);
  dfs(0, 1);
  for (i = 0; i < m; i++) cout << ans[i] << "\n";
}
