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 ssz[100005];
vector<int> g[100005];
int whichLvl[100005];
int pp[100005][20];
void getsz(int v, int p, int l) {
  ssz[v] = 1;
  whichLvl[v] = l;
  for (auto u : g[v])
    if (u != p) {
      getsz(u, v, l + 1);
      ssz[v] += ssz[u];
    }
}
vector<int> *vec[100005];
int cnt[100005];
int ans[100005];
vector<pair<int, int> > qq[100005];
void ans_query(int v) {
  for (int i = 0; i < qq[v].size(); i++) {
    int l = whichLvl[v] + qq[v][i].first;
    int idx = qq[v][i].second;
    if (l < 100005) ans[idx] = cnt[l] - 1;
  }
}
void dfs(int v, int p, bool keep) {
  int mx = -1, bigChild = -1;
  for (auto u : g[v])
    if (u != p && ssz[u] > mx) {
      mx = ssz[u];
      bigChild = u;
    }
  for (auto u : g[v])
    if (u != p && u != bigChild) dfs(u, v, 0);
  if (bigChild != -1) {
    dfs(bigChild, v, 1);
    vec[v] = vec[bigChild];
  } else
    vec[v] = new vector<int>();
  vec[v]->push_back(v);
  cnt[whichLvl[v]]++;
  for (auto u : g[v])
    if (u != p && u != bigChild)
      for (auto x : *vec[u]) {
        cnt[whichLvl[x]]++;
        vec[v]->push_back(x);
      }
  ans_query(v);
  if (keep == 0)
    for (auto u : *vec[v]) {
      cnt[whichLvl[u]]--;
    }
}
void par_calc() {
  for (int j = 1; j < 18; j++)
    for (int i = 0; i < 100005; i++)
      if (pp[i][j - 1] != -1) pp[i][j] = pp[pp[i][j - 1]][j - 1];
}
int get_par(int v, int n) {
  for (int i = 18; i >= 0; i--)
    if ((1 << i) <= n && pp[v][i] != -1) {
      v = pp[v][i];
      n -= 1 << i;
    }
  if (n || !v) return -1;
  return v;
}
int main() {
  int n, par, m, u, v;
  memset(pp, -1, sizeof(pp));
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &par);
    g[par].push_back(i);
    pp[i][0] = par;
  }
  par_calc();
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &u, &v);
    qq[get_par(u, v)].push_back(make_pair(v, i));
  }
  getsz(0, -1, 0);
  dfs(0, -1, 1);
  for (int i = 0; i < m; i++) printf("%d ", ans[i]);
  return 0;
}
