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 N = 1e5 + 9;
long long n, x, a[N], ans[N], m, v, k, dist[N], cnt, w[N], o, b1, b2, badi[N],
    l, r, mid, p, tvn = 262144, ct, l1 = 19, sum;
string s, b[N];
vector<long long> g[N], dis[N];
vector<pair<long long, long long> > u;
vector<vector<pair<long long, long long> > > q[N];
map<long long, long long> tedad[N];
map<string, long long> mk[N];
string binary(int m, int k) {
  string g = "";
  for (long long i = 0; i < k; i++) {
    g += char((m % 2) + '0');
    m = m / 2;
  }
  return g;
}
void dfs(int x, int ds) {
  badi[x] = N * 2;
  a[x] = cnt++, dist[x] = ds;
  if (dis[ds].size() > 0) badi[dis[ds][dis[ds].size() - 1]] = a[x];
  dis[ds].push_back(x);
  for (long long i = 0; i < g[x].size(); i++) {
    dfs(g[x][i], ds + 1);
  }
}
long long binary_search(long long x, long long d) {
  l = -1, r = dis[d].size();
  while (l + 1 < r) {
    mid = (l + r) / 2;
    if (a[dis[d][mid]] < x)
      l = mid;
    else
      r = mid;
  }
  return l;
}
int main() {
  scanf("%I64d", &n);
  ;
  for (long long i = 1; i < n + 1; i++) {
    cin >> b[i];
    scanf("%I64d", &w[i]);
    ;
    g[w[i]].push_back(i);
  }
  for (long long i = 1; i < n + 1; i++) {
    if (w[i] == 0) dfs(i, 0);
  }
  for (long long i = 0; i < n + 1; i++) {
    while (q[i].size() < dis[i].size()) q[i].push_back(u);
  }
  scanf("%I64d", &m);
  ;
  for (long long i = 1; i < m + 1; i++) {
    scanf("%I64d%I64d", &v, &k);
    ;
    o = dist[v] + k;
    if (o <= n + 1) {
      b1 = binary_search(a[v], o) + 1;
      b2 = binary_search(badi[v], o);
      ans[i] = b2 - b1 + 1;
      if (b2 >= b1) q[o][b2].push_back(make_pair(b1, i));
    }
  }
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < dis[i].size(); j++) {
      if (mk[i][b[dis[i][j]]] != 0) {
        p = 0, ct = tvn;
        s = binary(mk[i][b[dis[i][j]]], l1);
        for (int e = l1 - 1; e >= 0; e--) {
          if (s[e] == '1') {
            p += ct;
            tedad[i][p]++;
          }
          ct /= 2;
        }
      }
      mk[i][b[dis[i][j]]] = j + 1;
      for (long long z = 0; z < q[i][j].size(); z++) {
        sum = 0;
        s = binary(tvn * 2 - (q[i][j][z].first + 1), l1);
        p = tvn * 2, ct = tvn;
        for (int e = l1 - 1; e >= 0; e--) {
          if (s[e] == '1') {
            p -= ct;
            sum += tedad[i][p];
          }
          ct /= 2;
        }
        ans[q[i][j][z].second] -= sum;
      }
    }
  }
  for (long long i = 1; i < m + 1; i++) cout << ans[i] << " ";
}