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 N = 1e5 + 20, B = 20;
int n, a[N], q, v, p, str[N], fin[N], t, h[N], par[N][B], tmp1, tmp2;
vector<int> adj[N], dis[N];
void dfs(int u, int p) {
  str[u] = t;
  t++;
  for (int i = 1; i < B; i++) {
    if (par[u][i - 1] == -1) {
      break;
    }
    par[u][i] = par[par[u][i - 1]][i - 1];
  }
  for (auto x : adj[u]) {
    if (x != p) {
      h[x] = h[u] + 1;
      par[x][0] = u;
      dfs(x, u);
      dis[h[x]].push_back(str[x]);
    }
  }
  fin[u] = t;
}
inline int e_dis(int v, int d) {
  for (int i = 0; i < B; i++) {
    if (d & (1 << i)) {
      v = par[v][i];
    }
  }
  return v;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    adj[a[i]].push_back(i);
    adj[i].push_back(a[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0) {
      h[i] = 0;
      dis[h[i]].push_back(i);
      dfs(i, 0);
    }
  }
  cin >> q;
  while (q--) {
    cin >> v >> p;
    p = e_dis(v, p);
    if (p == 0) {
      cout << 0 << endl;
      continue;
    }
    tmp1 = lower_bound(dis[h[v]].begin(), dis[h[v]].end(), str[p]) -
           dis[h[v]].begin();
    tmp2 = lower_bound(dis[h[v]].begin(), dis[h[v]].end(), fin[p]) -
           dis[h[v]].begin();
    cout << (tmp2 - 1) - tmp1 + 1 - 1 << endl;
  }
  return 0;
}
