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 int MAX = 2 * 1001 * 100;
bool vis[MAX];
vector<int> a[MAX];
int st[MAX], en[MAX], h[MAX], cnt = 0;
vector<pair<int, int> > c[MAX];
vector<pair<int, int> > d[MAX];
void dfs(int v) {
  st[v] = cnt++;
  vis[v] = true;
  c[h[v]].push_back({st[v], v});
  for (auto u : a[v]) {
    if (!vis[u]) {
      h[u] = h[v] + 1;
      dfs(u);
    }
  }
  en[v] = cnt;
  d[h[v]].push_back({en[v], v});
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[x].push_back(i);
    a[i].push_back(x);
  }
  dfs(0);
  for (int x = 1; x <= n; x++) {
    sort(c[x].begin(), c[x].end());
    sort(d[x].begin(), d[x].end());
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, p;
    cin >> v >> p;
    int hp = h[v] - p;
    int H = h[v];
    if (hp <= 0) {
      cout << 0 << " ";
      continue;
    }
    int ind1 = lower_bound(c[hp].begin(), c[hp].end(),
                           pair<int, int>(st[v], -(1000 * 1000 * 10))) -
               c[hp].begin();
    ind1--;
    int newV = c[hp][ind1].second;
    int ans1 = lower_bound(c[H].begin(), c[H].end(),
                           pair<int, int>(st[newV], -(1000 * 1000 * 10))) -
               c[H].begin();
    int ans2 = lower_bound(c[H].begin(), c[H].end(),
                           pair<int, int>(en[newV], -(1000 * 1000 * 10))) -
               c[H].begin();
    cout << ans2 - ans1 - 1 << " ";
  }
}
