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 MAXN = 1e5 + 10;
int n, q, anc[MAXN][20], lev[MAXN], sz[MAXN], answer[MAXN];
vector<int> adj[MAXN], st, query[MAXN];
pair<int, int> qu[MAXN];
map<int, set<string>> mp[MAXN];
set<int> have[MAXN];
string s[MAXN];
void predfs(int v, int l) {
  lev[v] = l;
  sz[v] = 1;
  for (auto i : adj[v]) {
    predfs(i, l + 1);
    sz[v] += sz[i];
  }
}
void dfs(int v) {
  int mx = -1, ind = -1;
  for (auto i : adj[v]) {
    dfs(i);
    if (sz[i] > mx) {
      mx = sz[i];
      ind = i;
    }
  }
  if (ind != -1) {
    mp[ind].swap(mp[v]);
    have[ind].swap(have[v]);
  }
  mp[v][lev[v]].insert(s[v]);
  have[v].insert(lev[v]);
  for (auto i : adj[v]) {
    for (auto j : have[i]) {
      for (auto l : mp[i][j]) {
        mp[v][j].insert(l);
      }
      have[v].insert(j);
    }
    have[i].clear();
    mp[i].clear();
  }
  for (auto i : query[v]) {
    answer[i] = mp[v][qu[i].second].size();
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int v;
    cin >> s[i] >> v;
    if (v) {
      adj[v].push_back(i);
    } else {
      st.push_back(i);
    }
  }
  for (auto i : st) {
    predfs(i, 1);
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> v >> u;
    qu[i] = {v, lev[v] + u};
    query[v].push_back(i);
  }
  for (auto i : st) {
    dfs(i);
  }
  for (int i = 0; i < q; i++) {
    cout << answer[i] << '\n';
  }
  cin >> n;
  return 0;
}
