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;
int n, k, i, j, x, a, m, L[400000], R[400000], lavel;
vector<int> root, g[200005], s[200005];
int in[400000], timer, d[400000];
int p[20][100005], depth[100005];
void dfs(int v) {
  int i;
  L[v] = timer;
  in[v] = timer++;
  d[in[v]] = depth[v];
  for (i = 1; i < 20; i++) p[i][v] = p[i - 1][p[i - 1][v]];
  for (i = 0; i < g[v].size(); i++) {
    if (p[0][v] == g[v][i]) continue;
    depth[g[v][i]] = depth[v] + 1;
    p[0][g[v][i]] = v;
    dfs(g[v][i]);
  }
  R[v] = timer++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a;
    a--;
    if (a != -1)
      g[a].push_back(i);
    else
      root.push_back(i);
  }
  for (i = 0; i < root.size(); i++) dfs(root[i]);
  for (i = 0; i < timer; i++) s[d[i]].push_back(i);
  for (i = 0; i < timer; i++) sort(s[i].begin(), s[i].end());
  cin >> m;
  while (m--) {
    cin >> a >> lavel;
    a--;
    if (depth[a] < lavel) {
      cout << 0 << " ";
      continue;
    }
    x = depth[a];
    for (i = 19; i >= 0; i--) {
      if (lavel >= (1 << i)) {
        a = p[i][a];
        lavel -= (1 << i);
      }
    }
    cout << (upper_bound(s[x].begin(), s[x].end(), R[a]) -
             lower_bound(s[x].begin(), s[x].end(), L[a])) -
                1
         << " ";
  }
  return 0;
}
