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;
const int mxN = 1e5;
const int lvl = 17;
vector<int> cnt[mxN];
vector<int> adj[mxN];
int dpt[mxN];
int id[mxN];
int idx = 1;
int sub[mxN];
int n;
int prnt[mxN][lvl + 1];
int dfs(int u, int e, int d) {
  dpt[u] = d;
  sub[u] = 1;
  id[u] = idx++;
  cnt[dpt[u]].push_back(id[u]);
  prnt[u][0] = e;
  for (int i = 1; i <= lvl; i++) {
    if (prnt[u][i - 1] != -1) {
      prnt[u][i] = prnt[prnt[u][i - 1]][i - 1];
    }
  }
  for (auto ad = adj[u].begin(); ad != adj[u].end(); ad++) {
    sub[u] += dfs(*ad, u, d + 1);
  }
  return sub[u];
}
int calc(int u, int p) {
  int v = u;
  for (int i = lvl; i >= 0; i--) {
    if (p >= (1 << i)) {
      if (prnt[v][i] == -1) return 0;
      v = prnt[v][i];
      p -= 1 << i;
    }
  }
  auto mi = lower_bound(cnt[dpt[u]].begin(), cnt[dpt[u]].end(), id[v]);
  auto ma = lower_bound(cnt[dpt[u]].begin(), cnt[dpt[u]].end(), id[v] + sub[v]);
  ma--;
  return distance(mi, ma);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  vector<int> st;
  memset(prnt, -1, mxN * (lvl + 1) * 4);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    x--;
    if (x == -1)
      st.push_back(i);
    else
      adj[x].push_back(i);
  }
  for (int i = 0; i < st.size(); i++) {
    dfs(st[i], -1, 0);
  }
  int m;
  cin >> m;
  while (m--) {
    int v, p;
    cin >> v >> p;
    v--;
    cout << calc(v, p) << " ";
  }
  return 0;
}