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>
#pragma comment(linker, "/STACK:50000000")
using namespace std;
const int N = 100010;
int n, i, m, tim, r[N], v[N], p[N], d[N], tin[N], tout[N], a[N];
vector<int> g[N], h[N], sp[N];
map<pair<int, int>, int> f;
void dfs(int v, int dist) {
  int i;
  tin[v] = tim++;
  d[v] = dist;
  a[dist] = v;
  sp[dist].push_back(tin[v]);
  for (i = 0; i < (int)g[v].size(); i++) dfs(g[v][i], dist + 1);
  for (i = 0; i < (int)h[v].size(); i++)
    if (h[v][i] <= dist) f[make_pair(v, h[v][i])] = a[dist - h[v][i]];
  tout[v] = tim - 1;
}
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> r[i];
    if (r[i]) g[r[i] - 1].push_back(i);
  }
  cin >> m;
  for (i = 0; i < m; i++) {
    cin >> v[i] >> p[i];
    v[i]--;
    h[v[i]].push_back(p[i]);
  }
  tim = 0;
  for (i = 0; i < n; i++)
    if (!r[i]) dfs(i, 0);
  for (i = 0; i < m; i++) {
    if (p[i] > d[v[i]]) {
      cout << "0 ";
      continue;
    }
    int x = f[make_pair(v[i], p[i])], dist = d[v[i]], l = -1,
        r = (int)sp[dist].size(), mid;
    while (r - l > 1) {
      mid = (r + l) / 2;
      if (sp[dist][mid] < tin[x])
        l = mid;
      else
        r = mid;
    }
    int y = r;
    l = -1;
    r = (int)sp[dist].size();
    while (r - l > 1) {
      mid = (r + l) / 2;
      if (sp[dist][mid] > tout[x])
        r = mid;
      else
        l = mid;
    }
    cout << l - y << " ";
  }
  return 0;
}
