Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
long long n, q;
vector<long long> parent;
vector<vector<long long>> g;
vector<vector<long long>> jump;
vector<long long> tin;
vector<long long> tout;
vector<long long> level;
vector<long long> sub;
vector<long long> sorted = {};
long long timer = 0;
void dfs(long long v, long long p = 0) {
  if (tin[v] != -1) return;
  sorted.push_back(v);
  tin[v] = timer;
  timer++;
  if (v != 0) level[v] = level[p] + 1;
  for (long long to : g[v]) {
    dfs(to, v);
  }
  if (v != 0) sub[p] += sub[v];
  tout[v] = timer;
  timer++;
}
long long is_parent(long long u, long long v) {
  if (tin[v] <= tin[u] && tout[v] >= tout[u]) return v;
  if (tin[u] <= tin[v] && tout[u] >= tout[v]) return u;
  return -1;
}
bool strict_parent(long long u, long long v) {
  if (tin[u] <= tin[v] && tout[u] >= tout[v]) return true;
  return false;
}
long long lca(long long u, long long v) {
  if (u == v) return u;
  if (is_parent(u, v) != -1) return is_parent(u, v);
  for (long long x = 24; x >= 0; --x) {
    if (is_parent(jump[u][x], jump[v][x]) == -1) {
      u = jump[u][x];
      v = jump[v][x];
    }
  }
  return is_parent(parent[u], parent[v]);
}
long long do_jump(long long u, long long t) {
  if (t == 0) return u;
  for (long long x = 24; x >= 0; --x) {
    if (t >= (1 << x)) {
      u = jump[u][x];
      if (u == 0) return 0;
      u--;
      t -= (1 << x);
    }
  }
  return u + 1;
}
long long dist(long long u, long long v) {
  return level[u] + level[v] - 2 * level[lca(u, v)];
}
long long rate(long long s, long long f, long long t) {
  long long sf = lca(s, f);
  long long tf = lca(t, f);
  long long st = lca(s, t);
  long long ans = 0;
  if (strict_parent(sf, st)) ans += dist(sf, st);
  if (strict_parent(sf, tf)) ans += dist(tf, f);
  return ans;
}
vector<long long> order;
vector<bool> used;
void dfs1(long long v) {
  if (used[v]) return;
  used[v] = true;
  order.push_back(level[v]);
  for (long long to : g[v]) dfs1(to);
  order.push_back(level[v]);
}
struct counter {
  long long n;
  vector<long long> a;
  vector<vector<long long>> h;
  counter(vector<long long> _a) {
    n = _a.size();
    a = _a;
    h = vector<vector<long long>>(n, vector<long long>());
    for (long long i = 0; i < a.size(); ++i) {
      h[a[i]].push_back(i);
    }
  }
  long long count(long long x, long long l, long long r) {
    long long i = lower_bound(h[x].begin(), h[x].end(), l) - h[x].begin();
    long long j = lower_bound(h[x].begin(), h[x].end(), r) - h[x].begin();
    return j - i + 1;
  }
};
signed main() {
  cin >> n;
  parent = vector<long long>(n);
  g = vector<vector<long long>>(n, vector<long long>());
  jump = vector<vector<long long>>(n, vector<long long>(25, -1));
  tin = vector<long long>(n, -1);
  tout = vector<long long>(n, -1);
  level = vector<long long>(n, 0);
  sub = vector<long long>(n, 1);
  order = vector<long long>();
  used = vector<bool>(n, false);
  long long p;
  for (long long i = 0; i < n; ++i) {
    long long v = i;
    long long u;
    cin >> u;
    parent[i] = u;
    if (u == 0) continue;
    u--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (long long v = 0; v < n; ++v)
    if (parent[v] == 0) dfs(v);
  for (long long v = 0; v < n; ++v)
    if (parent[v] == 0) dfs1(v);
  for (long long i = 0; i < n; ++i) {
    jump[i][0] = parent[i];
  }
  for (long long u = 0; u < n; ++u) {
    long long i = sorted[u];
    for (long long j = 1; j < 25; ++j) {
      if (jump[i][j - 1] != 0)
        jump[i][j] = jump[jump[i][j - 1] - 1][j - 1];
      else
        jump[i][j] = 0;
    }
  }
  long long m;
  cin >> m;
  counter cntr(order);
  for (long long i = 0; i < m; ++i) {
    long long v, p;
    cin >> v >> p;
    v--;
    long long pred = do_jump(v, p);
    if (pred == 0) {
      cout << 0 << " ";
      continue;
    }
    pred--;
    long long ans = -2;
    ans += cntr.count(level[v], tin[pred], tout[pred]);
    ans = ans / 2;
    cout << ans << " ";
  }
  cout << endl;
  return 0;
}