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;
char name[100001][30];
int n, m, mark[100001][2], depth[100001], step;
vector<int> graph[100001], level[100001];
map<pair<int, int>, int> cache;
void dfs(int u, int d) {
  mark[u][0] = step++;
  depth[u] = d;
  level[d].push_back(u);
  for (int i = 0, sz = graph[u].size(); i < sz; i++) dfs(graph[u][i], d + 1);
  mark[u][1] = step++;
}
inline int bs(vector<int> &v, int c) {
  int i = 0, j = v.size();
  if (j == 0) return 0;
  while (i + 1 < j) {
    int m = (i + j) / 2;
    if (mark[v[m]][0] <= c)
      i = m;
    else
      j = m;
  }
  if (mark[v[i]][0] > c) return i;
  return j;
}
inline int ksons(int v, int k) {
  if (depth[v] + k > n) return 0;
  if (cache.count(make_pair(v, k))) return cache[make_pair(v, k)];
  set<string> s;
  vector<int> &ve = level[depth[v] + k];
  int tot = 0;
  for (int x = bs(ve, mark[v][0]), sz = ve.size();
       x < sz && mark[ve[x]][1] < mark[v][1]; x++) {
    s.insert(name[ve[x]]);
  }
  return cache[make_pair(v, k)] = s.size();
}
int main() {
  int a, v, k;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s %d", &name[i][0], &a);
    graph[a].push_back(i);
  }
  dfs(0, 0);
  scanf("%d", &m);
  while (m--) {
    scanf("%d %d", &v, &k);
    printf("%d\n", ksons(v, k));
  }
  return 0;
}
