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;
vector<int> MAP[100005];
vector<int> deep[100005];
int in[100005];
int out[100005];
int who[100005];
int dep[100005];
int id;
map<int, string> STR;
map<string, int> ID;
int n, m;
map<pair<int, int>, int> ans;
void dfs(int u, int d) {
  in[u] = ++id;
  dep[u] = d;
  who[id] = u;
  for (unsigned int i = 0; i < MAP[u].size(); i++) dfs(MAP[u][i], d + 1);
  out[u] = id;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    string str;
    int tmp;
    cin >> str >> tmp;
    STR[i] = str;
    ID[str] = tmp;
    MAP[tmp].push_back(i);
  }
  dfs(0, 0);
  for (int i = 0; i <= n; i++) deep[dep[i]].push_back(in[i]);
  for (int i = 0; i <= n; i++) sort(deep[i].begin(), deep[i].end());
  scanf("%d", &m);
  while (m--) {
    int v, k;
    scanf("%d %d", &v, &k);
    int d = dep[v] + k;
    if (d > 100000) {
      printf("0\n");
      continue;
    }
    if (ans[make_pair(v, k)]) {
      printf("%d\n", ans[make_pair(v, k)]);
      continue;
    }
    int k1 =
        lower_bound(deep[d].begin(), deep[d].end(), in[v]) - deep[d].begin();
    int k2 =
        upper_bound(deep[d].begin(), deep[d].end(), out[v]) - deep[d].begin();
    k2--;
    set<string> SET;
    for (int i = k1; i <= k2; i++) {
      SET.insert(STR[who[deep[d][i]]]);
    }
    printf("%d\n", SET.size());
    ans[make_pair(v, k)] = SET.size();
  }
  return 0;
}
