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>
inline int sbt(int x) { return __builtin_popcount(x); }
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
vector<int> adj[100013];
int par[100013][33];
vector<int> dfslevel[100013];
int start[100013];
int endt[100013];
int lev[100013];
int n, k;
int timer = 0;
void dfs(int node, int p, int level) {
  start[node] = timer;
  timer++;
  lev[node] = level;
  dfslevel[level].push_back(start[node]);
  par[node][0] = p;
  for (auto nei : adj[node]) {
    if (nei != p) {
      dfs(nei, node, level + 1);
    }
  }
  endt[node] = timer;
  timer++;
}
int f(int node) {
  int curr = node;
  for (int i = 33 - 1; i >= 0; i--) {
    if (k & (1ll << i)) {
      curr = par[curr][i];
    }
  }
  return curr;
}
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  dfs(0, -1, 0);
  for (int i = 1; i < 33; i++) {
    for (int node = 1; node <= n; node++) {
      par[node][i] = par[par[node][i - 1]][i - 1];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int u;
    cin >> u >> k;
    int node_level = lev[u];
    int target = f(u);
    if (target <= 0) {
      cout << 0 << ' ';
      continue;
    }
    int shuru = start[target];
    int khatam = endt[target];
    int same_lev = (lower_bound(dfslevel[node_level].begin(),
                                dfslevel[node_level].end(), khatam) -
                    lower_bound(dfslevel[node_level].begin(),
                                dfslevel[node_level].end(), shuru)) -
                   1;
    cout << same_lev << ' ';
  }
  return 0;
}
