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;
int const N = 1e5 + 5;
vector<vector<int>> adj(N);
int indexToNode[N], NodeToIndex[N], in[N], out[N], sz[N], depth[N];
int cnt = 1;
void dfs(int u, int prt) {
  indexToNode[cnt] = u;
  NodeToIndex[u] = cnt;
  sz[u] = 1;
  in[u] = cnt++;
  for (int v : adj[u]) {
    if (v == prt) continue;
    depth[v] = depth[u] + 1;
    dfs(v, u);
    sz[u] += sz[v];
  }
  out[u] = cnt - 1;
}
string s[N];
vector<vector<pair<int, int>>> quer(N);
int ans[N];
deque<set<string>> smallToLarge(int u, int prt) {
  deque<set<string>> ret;
  int big = -1;
  for (int v : adj[u]) {
    if (v == prt) continue;
    if (sz[v] * 2 > sz[u]) {
      big = v;
    } else {
      smallToLarge(v, u);
    }
  }
  if (big != -1) ret = smallToLarge(big, u);
  ret.emplace_front(set<string>({s[u]}));
  for (int v : adj[u]) {
    if (v == prt || v == big) continue;
    for (int i = in[v]; i <= out[v]; i++) {
      int w = indexToNode[i];
      int idx = depth[w] - depth[u];
      if (ret.size() <= idx) ret.resize(idx + 1);
      ret[idx].emplace(s[w]);
    }
  }
  for (auto &[k, i] : quer[u]) {
    if (k >= ret.size())
      ans[i] = 0;
    else
      ans[i] = ret[k].size();
  }
  return ret;
}
void test() {
  int n;
  cin >> n;
  vector<int> roots;
  for (int i = 1; i <= n; i++) {
    string temp;
    int r;
    cin >> temp >> r;
    s[i] = temp;
    if (r == 0)
      roots.push_back(i);
    else
      adj[r].push_back(i);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, k;
    cin >> v >> k;
    quer[v].emplace_back(k, i);
  }
  for (int i : roots) dfs(i, 0), smallToLarge(i, 0);
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    test();
  }
}
