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;
long long n, m, k = 0, ans = 0, maxh;
vector<int> a[(1000 * 1000 + 7)];
vector<pair<pair<int, int>, int> > b[(1000 * 1000 + 7)];
int h[(1000 * 1000 + 7)], s[(1000 * 1000 + 7)], e[(1000 * 1000 + 7)];
bool mark[(1000 * 1000 + 7)];
bool cmp1(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.first < b.first.first;
}
bool cmp2(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.second < b.first.second;
}
int dfs(int v, int t) {
  mark[v] = true;
  int st = s[v];
  for (int i = 0; i < a[v].size(); i++) {
    int u = a[v][i];
    s[u] = st + 1;
    h[u] = t + 1;
    st = dfs(u, t + 1);
  }
  maxh = max(maxh, (long long)t);
  e[v] = st;
  b[t].push_back(make_pair(make_pair(s[v], e[v]), v));
  return st + 1;
}
void solve(int v, int p) {
  if (h[v] <= p) {
    cout << 0 << " ";
    return;
  }
  if (p == 0) {
    cout << 0 << " ";
    return;
  }
  vector<pair<pair<int, int>, int> >::iterator u =
      lower_bound(b[h[v] - p].begin(), b[h[v] - p].end(),
                  make_pair(make_pair(s[v], e[v]), v), cmp1);
  u--;
  v = u->second;
  vector<pair<pair<int, int>, int> >::iterator i =
      lower_bound(b[h[v] + p].begin(), b[h[v] + p].end(),
                  make_pair(make_pair(s[v], e[v]), v), cmp1);
  vector<pair<pair<int, int>, int> >::iterator j =
      lower_bound(b[h[v] + p].begin(), b[h[v] + p].end(),
                  make_pair(make_pair(s[v], e[v]), v), cmp2);
  j--;
  if (j->second > n || i->second > n || j->second < 0 || i->second < 0) {
    cout << 0 << " ";
    return;
  }
  cout << j - i << " ";
}
int main() {
  cin.tie();
  cout.tie();
  ios_base::sync_with_stdio(false);
  cin >> n;
  int x, y;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    a[x].push_back(i);
  }
  dfs(0, 0);
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    solve(x, y);
  }
}
