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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = 1000 * 1000 * 1000;
const int MXN = 100100 << 1;
struct Nd {
  int sum;
  Nd *l, *r;
  Nd(int _sum = 0) {
    sum = _sum;
    l = r = 0;
  }
};
typedef Nd* tree;
int hoho(tree t) {
  if (t == 0) {
    return 0;
  }
  return t->sum;
}
tree build(int tl, int tr) {
  if (tl == tr) {
    return new Nd();
  }
  int tm = (tl + tr) >> 1;
  tree v = new Nd();
  v->l = build(tl, tm);
  v->r = build(tm + 1, tr);
  v->sum = hoho(v->l) + hoho(v->r);
  return v;
}
int getSum(int l, int r, tree t, int tl, int tr) {
  if (l == tl && r == tr) {
    return t->sum;
  }
  int tm = (tl + tr) >> 1;
  if (r <= tm) {
    return getSum(l, r, t->l, tl, tm);
  }
  if (l > tm) {
    return getSum(l, r, t->r, tm + 1, tr);
  }
  return getSum(l, tm, t->l, tl, tm) + getSum(tm + 1, r, t->r, tm + 1, tr);
}
tree update(int pos, int nval, tree t, int tl, int tr) {
  if (tl == tr) {
    return new Nd(nval);
  }
  int tm = (tl + tr) >> 1;
  tree v = new Nd();
  v->l = t->l;
  v->r = t->r;
  if (pos <= tm) {
    v->l = update(pos, nval, t->l, tl, tm);
  } else {
    v->r = update(pos, nval, t->r, tm + 1, tr);
  }
  v->sum = hoho(v->l) + hoho(v->r);
  return v;
}
vector<vector<int> > g;
vector<int> roots;
vector<tree> trees[MXN];
vector<int> h[MXN];
int pizda[MXN];
string tos[MXN];
int d[MXN];
int q[MXN];
map<string, int> toi;
int timer;
int dfs(int v, int hui = -1, int hgt = 0) {
  pizda[v] = timer++;
  d[v] = hgt;
  q[v] = 1;
  h[hgt].push_back(v);
  for (int i = (0); i < ((int)g[v].size()); ++i) {
    int to = g[v][i];
    if (to != hui) {
      q[v] += dfs(to, v, hgt + 1);
    }
  }
  return q[v];
}
void prepare(int hgt) {
  map<int, int> ps;
  tree curTree = build(0, (int)h[hgt].size() - 1);
  for (int i = (0); i < ((int)h[hgt].size()); ++i) {
    int tm = toi[tos[h[hgt][i]]];
    if (ps.find(tm) == ps.end()) {
      ps.insert(make_pair(tm, i));
      curTree = update(i, 1, curTree, 0, (int)h[hgt].size() - 1);
    } else {
      curTree = update(ps[tm], 0, curTree, 0, (int)h[hgt].size() - 1);
      ps[tm] = i;
      curTree = update(i, 1, curTree, 0, (int)h[hgt].size() - 1);
    }
    trees[hgt].push_back(curTree);
  }
}
int bsearch(int hgt, int val) {
  int l = 0, r = (int)h[hgt].size() - 1;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (pizda[h[hgt][mid]] < val) {
      l = mid + 1;
    } else {
      r = mid;
    }
  }
  return l;
}
int query(int v, int k) {
  int hgt = d[v] + k;
  if ((h[hgt]).empty()) {
    return 0;
  }
  if (pizda[v] + 1 > pizda[h[hgt].back()] ||
      pizda[v] + q[v] < pizda[h[hgt].front()]) {
    return 0;
  }
  int l = bsearch(hgt, pizda[v] + 1), r = bsearch(hgt, pizda[v] + q[v]);
  while (r >= 0 && pizda[h[hgt][r]] > pizda[v] + q[v]) {
    --r;
  }
  if (l > r) {
    return 0;
  }
  return getSum(l, r, trees[hgt][r], 0, (int)h[hgt].size() - 1);
}
int main() {
  int N;
  scanf("%d", &N);
  g.resize(N);
  for (int i = (0); i < (N); ++i) {
    string name;
    cin >> name;
    int hui;
    scanf("%d", &hui);
    if (hui == 0) {
      roots.push_back(i);
    } else {
      g[hui - 1].push_back(i);
    }
    int t = (int)toi.size();
    toi.insert(make_pair(name, t));
    tos[i] = name;
  }
  for (int i = (0); i < ((int)roots.size()); ++i) {
    dfs(roots[i]);
  }
  for (int i = (0); i < (N); ++i)
    if ((int)h[i].size()) {
      prepare(i);
    }
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int v, k;
    scanf("%d%d", &v, &k);
    printf("%d\n", query(v - 1, k));
  }
  return 0;
}
