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 BC[200005], sz[200005], res[200005], n, m, clr[200005], cnt[200005];
bool mark[200005];
multiset<int> s[200005];
map<string, int> mp;
vector<int> graph[200005];
vector<int> par;
vector<pair<int, int> > qur[200005];
multiset<int>::iterator it;
void getsz(int u, int p) {
  int best = -1;
  BC[u] = -1;
  sz[u] = 1;
  int len = graph[u].size();
  for (int i = 0; i < len; i++) {
    int v = graph[u][i];
    if (v == p) continue;
    getsz(v, u);
    sz[u] += sz[v];
    if (sz[v] > best) {
      best = sz[v];
      BC[u] = v;
    }
  }
}
void add(int u, int p, int level, int x) {
  if (x == 1) {
    if (s[level].find(clr[u]) == s[level].end()) cnt[level]++;
    s[level].insert(clr[u]);
  }
  if (x == -1) {
    it = s[level].find(clr[u]);
    if (it != s[level].end()) {
      s[level].erase(it);
      if (s[level].find(clr[u]) == s[level].end()) cnt[level]--;
    }
  }
  int len = graph[u].size();
  for (int i = 0; i < len; i++) {
    int v = graph[u][i];
    if (v != p and mark[v] == 0) add(v, u, level + 1, x);
  }
}
void dfs(int u, int p, int level, int clean) {
  int len = graph[u].size();
  for (int i = 0; i < len; i++) {
    int v = graph[u][i];
    if (v != p and BC[u] != v) dfs(v, u, level + 1, 0);
  }
  if (BC[u] != -1) {
    dfs(BC[u], u, level + 1, 1);
    mark[BC[u]] = 1;
  }
  add(u, p, level, 1);
  int len1 = qur[u].size();
  for (int i = 0; i < len1; i++) {
    pair<int, int> temp = qur[u][i];
    res[temp.second] = cnt[level + temp.first];
  }
  if (BC[u] != -1) mark[BC[u]] = 0;
  if (clean == 0) {
    add(u, p, level, -1);
  }
}
int main() {
  string str;
  cin >> n;
  int k = 0;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> str >> p;
    if (!mp[str]) {
      k++;
      mp[str] = k;
    }
    clr[i + 1] = mp[str];
    graph[p].push_back(i + 1);
    graph[i + 1].push_back(p);
    if (p == 0) par.push_back(i + 1);
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, k;
    scanf("%d%d", &v, &k);
    qur[v].push_back(pair<int, int>(k, i));
  }
  int len = par.size();
  for (int i = 0; i < len; i++) {
    getsz(par[i], 0);
    dfs(par[i], 0, 1, 0);
  }
  for (int i = 0; i < m; i++) printf("%d\n", res[i]);
  return 0;
}
