Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void my_dbg() { cout << endl; }
template <typename Arg, typename... Args>
void my_dbg(Arg A, Args... B) {
  cout << ' ' << A;
  my_dbg(B...);
}
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
const int inf = INT_MAX;
const int ninf = INT_MIN;
const int mod = 1e9 + 7;
const int maxN = 1e6 + 2;
vector<string> val(maxN);
set<string> total[maxN];
vector<pair<int, int>> q[maxN];
vi in(maxN), out(maxN), ver(maxN), sz(maxN), lev(maxN), ans(maxN);
vi adj[maxN];
int tim;
void dfs(int node, int par) {
  in[node] = tim;
  ver[tim] = node;
  sz[node] = 1;
  for (int it : adj[node]) {
    if (it != par) {
      tim++;
      lev[it] = lev[node] + 1;
      dfs(it, node);
      sz[node] += sz[it];
    }
  }
  out[node] = tim;
}
void dsu_on_tree(int node, int par, bool keep) {
  int mx = -1, big_child = -1;
  for (int it : adj[node]) {
    if (it != par && sz[it] > mx) {
      mx = sz[it];
      big_child = it;
    }
  }
  for (int it : adj[node]) {
    if (it != par && it != big_child) dsu_on_tree(it, node, 0);
  }
  if (big_child != -1) dsu_on_tree(big_child, node, 1);
  total[lev[node]].insert(val[node]);
  for (int it : adj[node]) {
    if (it != par && it != big_child) {
      for (int t = in[it]; t <= out[it]; t++) {
        int u = ver[t];
        total[lev[u]].insert(val[u]);
      }
    }
  }
  for (auto query : q[node]) {
    int depth = lev[node] + query.second, idx = query.first;
    if (depth < maxN)
      ans[idx] = (int)total[depth].size();
    else
      ans[idx] = 0;
  }
  if (!keep) {
    for (int t = in[node]; t <= out[node]; t++) {
      int u = ver[t];
      total[lev[u]].erase(val[u]);
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  vi roots;
  for (int i = 1; i < (int)n + 1; i++) {
    int p;
    cin >> val[i] >> p;
    if (p != 0) {
      adj[i].push_back(p);
      adj[p].push_back(i);
    } else {
      roots.push_back(i);
    }
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < (int)m; i++) {
    int v, k;
    scanf("%d", &v);
    scanf("%d", &k);
    q[v].push_back(make_pair(i, k));
  }
  for (int it : roots) {
    tim++;
    dfs(it, -1);
  }
  for (int it : roots) {
    dsu_on_tree(it, -1, 0);
  }
  for (int i = 0; i < (int)m; i++) {
    printf("%d ", (int)(ans[i]));
  }
  return 0;
}