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 logn = 17;
const int maxn = 100000;
const int ts = 1 << 17;
vector<int> g[maxn];
int tin[maxn], tout[maxn];
string str[maxn];
unordered_map<string, int> bank;
int n, _time = 0;
int ver[maxn], d[maxn];
int color[maxn];
long long ans[maxn];
unordered_set<int> cnt[maxn];
int sz[maxn];
struct query {
  int p, i;
  query() {}
  query(int _p, int _i) {
    p = _p;
    i = _i;
  }
};
vector<query> q[maxn];
void dfs_sz(int v) {
  ver[_time] = v;
  tin[v] = _time++;
  sz[v] = 1;
  for (auto u : g[v]) {
    d[u] = d[v] + 1;
    dfs_sz(u);
    sz[v] += sz[u];
  }
  tout[v] = _time;
}
void dfs(int v, bool keep = 0) {
  int big = -1;
  for (auto u : g[v]) {
    if ((big == -1 || sz[u] > sz[big])) big = u;
  }
  for (auto u : g[v]) {
    if (u != big) dfs(u, 0);
  }
  if (big != -1) dfs(big, 1);
  cnt[d[v]].insert(color[v]);
  for (auto u : g[v]) {
    if (u != big)
      for (int i = tin[u]; i < tout[u]; ++i)
        cnt[d[ver[i]]].insert(color[ver[i]]);
  }
  for (auto x : q[v]) {
    if (d[v] + x.p < n) ans[x.i] = cnt[d[v] + x.p].size();
  }
  if (!keep)
    for (int i = tin[v]; i < tout[v]; ++i) cnt[d[ver[i]]].clear();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<int> root;
  for (int i = 0, sz = 1; i < n; ++i) {
    int v;
    string name;
    cin >> name >> v;
    if (bank[name] != 0)
      color[i] = bank[name];
    else {
      bank[name] = sz;
      color[i] = sz++;
    }
    if (v)
      g[v - 1].push_back(i);
    else
      root.push_back(i);
  }
  for (auto v : root) dfs_sz(v);
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int v, p;
    cin >> v >> p;
    q[v - 1].push_back(query(p, i));
  }
  for (auto v : root) dfs(v);
  for (int i = 0; i < m; ++i) cout << ans[i] << '\n';
  return 0;
}
