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 pr[MAXN];
const int p = 31;
const int mod = 1e9 + 9;
int deg[40];
class Hech {
 public:
  long long h;
  Hech(){};
  Hech(long long h) : h(h){};
  void Get_Hech(string &s) {
    long long ans = 0;
    for (int i = 0; i < s.size(); i++) ans = (ans * p + s[i] - 'a' + 1) % mod;
    h = ans;
  }
};
struct node {
  node *l, *r;
  int cnt, y;
  pair<int, int> x;
  node(pair<int, int> x, int cnt) : x(x), cnt(cnt) {
    y = (rand() << 16) + rand();
    l = r = NULL;
  };
  node() {
    l = r = NULL;
    y = (rand() << 16) + rand();
    cnt = 0;
  }
};
typedef node *pnode;
class Tree {
 public:
  pnode root;
  int plus;
  void Build() {
    plus = 0;
    root = NULL;
  }
  int Cnt(pnode root) { return ((!root) ? 0 : (root->cnt)); }
  void Up(pnode root) { root->cnt = Cnt(root->l) + Cnt(root->r) + 1; }
  pnode Merge(pnode l, pnode r) {
    if (!l) return r;
    if (!r) return l;
    if (l->y > r->y) {
      l->r = Merge(l->r, r);
      Up(l);
      return l;
    } else {
      r->l = Merge(l, r->l);
      Up(r);
      return r;
    }
  }
  void Split(pnode root, pnode &l, pnode &r, pair<int, int> x) {
    if (!root) {
      l = r = NULL;
      return;
    }
    if ((root->x) <= x) {
      Split(root->r, root->r, r, x);
      Up(root);
      l = root;
    } else {
      Split(root->l, l, root->l, x);
      Up(root);
      r = root;
    }
  }
  bool Find(pnode root, int x, int idn) {
    if (!root) return 0;
    if ((root->x) == make_pair(x, idn)) return 1;
    if (root->x > make_pair(x, idn)) return Find(root->l, x, idn);
    return Find(root->r, x, idn);
  }
  void Insert(int x, int idn) {
    x -= plus;
    if (!Find(root, x, idn)) {
      pnode fi, se;
      Split(root, fi, se, make_pair(x, idn));
      root = Merge(Merge(fi, new node(make_pair(x, idn), 1)), se);
    }
  }
  int Cnt(int x) {
    x = x - plus;
    pnode fi, se, th;
    Split(root, se, th, make_pair(x + 1, -1));
    Split(se, fi, se, make_pair(x, -1));
    int ans = Cnt(se);
    root = Merge(Merge(fi, se), th);
    return ans;
  }
  int size() { return Cnt(root); }
};
Hech name[MAXN];
vector<int> g[MAXN];
int ans[MAXN];
vector<pair<int, int> > request[MAXN];
vector<int> root;
void DFS(pnode a, int plus, Tree &b) {
  if (!a) return;
  DFS(a->l, plus, b);
  DFS(a->r, plus, b);
  b.Insert(a->x.first + plus, a->x.second);
}
Tree Merge(Tree a, Tree b) {
  if (a.size() > b.size()) swap(a, b);
  DFS(a.root, a.plus, b);
  return b;
}
Tree DFS(int v) {
  Tree a;
  a.Build();
  a.Insert(0, name[v].h);
  for (int i = 0; i < g[v].size(); i++) {
    Tree ne = DFS(g[v][i]);
    ne.plus++;
    a = Merge(a, ne);
  }
  for (int i = 0; i < request[v].size(); i++) {
    int h = request[v][i].first;
    ans[request[v][i].second] = a.Cnt(h);
  }
  return a;
}
int main() {
  int n;
  scanf("%i", &n);
  string s;
  for (int i = 0; i < n; i++) {
    cin >> ws;
    cin >> s;
    scanf("%i", &pr[i]);
    name[i].Get_Hech(s);
    pr[i]--;
    if (pr[i] >= 0)
      g[pr[i]].push_back(i);
    else
      root.push_back(i);
  }
  int m;
  scanf("%i", &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%i %i", &a, &b);
    request[a - 1].push_back(make_pair(b, i));
  }
  for (int i = 0; i < root.size(); i++) DFS(root[i]);
  for (int i = 0; i < m; i++) printf("%i\n", ans[i]);
  return 0;
}
