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 ym[4] = {0, 0, -1, 1};
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
const long long POW = 9973;
string ar[MAXN];
vector<int> adj[MAXN];
int ht[MAXN];
map<string, int> hsh;
deque<unordered_set<int>> m[MAXN];
vector<vector<pair<int, int>>> qry(MAXN);
int vis[MAXN];
int ans[MAXN];
void dfs(int x) {
  vis[x] = 1;
  for (int i : adj[x]) {
    ht[i] = ht[x] + 1;
    dfs(i);
    if (m[i].size() > m[x].size()) swap(m[i], m[x]);
    for (int k = 0; k < m[i].size(); k++) {
      if (k >= m[x].size()) {
        m[x].push_back(m[i][k]);
        continue;
      }
      if (m[i][k].size() > m[x][k].size()) swap(m[x][k], m[i][k]);
      for (auto t : m[i][k]) m[x][k].insert(t);
    }
    m[i].clear();
  }
  unordered_set<int> next;
  next.insert(hsh[ar[x]]);
  m[x].push_front(next);
  for (auto p : qry[x]) {
    if (p.first >= m[x].size()) continue;
    ans[p.second] = m[x][p.first].size();
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    hsh[ar[i]] = i;
    int v;
    cin >> v;
    if (v) adj[v - 1].push_back(i);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int v, k;
    cin >> v >> k;
    v--;
    qry[v].push_back(make_pair(k, i));
  }
  for (int i = 0; i < n; i++) {
    if (!vis[i]) dfs(i);
  }
  for (int i = 0; i < q; i++) cout << ans[i] << "\n";
}
