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 long long INF = 1e9 + 10, MAX = 2e5 + 1e4, MOD = 1e9 + 7, MAXL = 25;
void OUT(long double o, int x) {
  cout << fixed << setprecision(x) << o;
  return;
}
long long dp[MAX][MAXL], h[MAX], t = 0, st[MAX], fi[MAX];
bool vis[MAX];
vector<pair<long long, long long> > vh[MAX];
vector<int> adj[MAX];
void dfs(int v, int p) {
  vis[v] = 1;
  dp[v][0] = p;
  h[v] = h[p] + 1;
  st[v] = t;
  t++;
  for (int i = 1; i <= 20; i++) dp[v][i] = dp[dp[v][i - 1]][i - 1];
  for (int u : adj[v])
    if (u != p) dfs(u, v);
  fi[v] = t;
  vh[h[v]].push_back({st[v], t});
  return;
}
int getp(int v, int k) {
  for (int i = 0; i <= 20; i++)
    if (k & (1 << i)) v = dp[v][i];
  return v;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    if (x != 0) {
      adj[i].push_back(x);
      adj[x].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, i);
  }
  for (int i = 1; i <= n; i++) sort(vh[i].begin(), vh[i].end());
  cin >> m;
  while (m--) {
    long long x, y;
    cin >> x >> y;
    if (h[x] <= y)
      cout << "0 ";
    else {
      long long v = getp(x, y), ans = 0;
      pair<long long, long long> p1 = {fi[v] + 1, 0}, p2 = {st[v], 0};
      ans =
          lower_bound(vh[h[x]].begin(), vh[h[x]].end(), p1) - vh[h[x]].begin();
      ans -=
          lower_bound(vh[h[x]].begin(), vh[h[x]].end(), p2) - vh[h[x]].begin();
      cout << ans - 1 << " ";
    }
  }
  return 0;
}
