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;
vector<vector<int> > adj, level;
int par[100001];
int in[100001], out[100001], rnk[100001];
int timer, n;
void dfs(int i, int depth) {
  in[i] = timer++;
  rnk[i] = depth;
  level[depth].push_back(in[i]);
  for (int j = 0; j < adj[i].size(); j++) {
    int next = adj[i][j];
    dfs(next, depth + 1);
  }
  out[i] = timer++;
}
int sparse[100001][32];
void build() {
  for (int i = 0; i < n; i++) {
    sparse[i][0] = par[i];
  }
  for (int i = 1; i < 32; i++) {
    for (int j = 0; j < n; j++) {
      if (sparse[j][i - 1] != -1) {
        sparse[j][i] = sparse[sparse[j][i - 1]][i - 1];
      }
    }
  }
}
int pth(int node, int p) {
  int tmp = 30;
  while (p > 0 && node >= 0) {
    while ((1 << tmp) > p) {
      tmp--;
    }
    node = sparse[node][tmp];
    p -= (1 << tmp);
  }
  return node;
}
int main() {
  while (cin >> n) {
    timer = 0;
    memset(par, -1, sizeof(par));
    memset(sparse, -1, sizeof(sparse));
    adj.clear();
    adj.resize(n);
    level.clear();
    level.resize(n);
    int u;
    for (int i = 0; i < n; i++) {
      cin >> u;
      if (u) {
        par[i] = u - 1;
        adj[u - 1].push_back(i);
      }
    }
    for (int i = 0; i < n; i++) {
      if (par[i] == -1) {
        dfs(i, 0);
      }
    }
    for (int i = 0; i < n; i++) {
      sort(level[i].begin(), level[i].end());
    }
    build();
    int m;
    cin >> m;
    int p;
    while (m--) {
      cin >> u >> p;
      u--;
      int parent = pth(u, p);
      if (parent == -1) {
        cout << 0 << endl;
      } else {
        cout << lower_bound(level[rnk[u]].begin(), level[rnk[u]].end(),
                            out[parent]) -
                    lower_bound(level[rnk[u]].begin(), level[rnk[u]].end(),
                                in[parent]) -
                    1
             << endl;
      }
    }
  }
  return 0;
}