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 N = 300010;
unordered_map<string, int> m[N];
int sz[N], bigchild[N], lev[N], ans[N];
vector<int> adj[N];
vector<pair<int, int> > q[N];
string s[N];
void compute(int i, int p) {
  sz[i] = 1;
  for (auto ve : adj[i]) {
    lev[ve] = lev[i] + 1;
    compute(ve, i);
    sz[i] += sz[ve];
  }
}
void add(int i, int p) {
  m[lev[i]][s[i]]++;
  for (auto ve : adj[i])
    if (!bigchild[ve]) add(ve, i);
}
void rem(int i, int p) {
  auto it = m[lev[i]].find(s[i]);
  it->second--;
  if (!it->second) m[lev[i]].erase(s[i]);
  for (auto ve : adj[i])
    if (!bigchild[ve]) rem(ve, i);
}
void dfs(int i, int p, int keep) {
  int mx = 0, sc = -1;
  for (auto ve : adj[i])
    if (sz[ve] > mx) {
      mx = sz[ve];
      sc = ve;
    }
  for (auto ve : adj[i]) {
    if (ve - sc) dfs(ve, i, 1);
  }
  if (sc + 1) {
    dfs(sc, i, 0);
    bigchild[sc] = 1;
  }
  add(i, p);
  if (sc + 1) {
    bigchild[sc] = 0;
  }
  for (auto query : q[i]) {
    ans[query.second] = m[lev[i] + query.first].size();
  }
  if (keep) rem(i, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t, n, i, j, k, r, root, v;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> s[i] >> r;
    adj[r].push_back(i);
  }
  root = 0;
  cin >> r;
  for (i = 0; i < r; i++) {
    cin >> v >> k;
    q[v].push_back(make_pair(k, i));
  }
  compute(root, -1);
  dfs(root, -1, 0);
  for (i = 0; i < r; i++) cout << ans[i] << "\n";
  return 0;
}
