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> adj[100010];
vector<pair<int, int> > point[100010];
map<pair<int, int>, int> vis;
int n, v, k, depth[100010], pos[100010][2], tot;
char s[100010][23];
void Dfs(int k, int d) {
  pos[k][0] = ++tot;
  point[d].push_back(make_pair(tot, k));
  depth[k] = d;
  for (vector<int>::iterator p = adj[k].begin(); p != adj[k].end(); ++p) {
    Dfs(*p, d + 1);
  }
  pos[k][1] = ++tot;
}
int solve() {
  if (vis[make_pair(v, k)]) return vis[make_pair(v, k)] - 1;
  vector<pair<int, int> >::iterator l, r;
  set<string> str;
  l = lower_bound(point[depth[v] + k].begin(), point[depth[v] + k].end(),
                  make_pair(pos[v][0], 100010));
  r = lower_bound(point[depth[v] + k].begin(), point[depth[v] + k].end(),
                  make_pair(pos[v][1], 0));
  for (vector<pair<int, int> >::iterator p = l; p != r; ++p) {
    str.insert(s[(*p).second]);
  }
  return (vis[make_pair(v, k)] = str.size() + 1) - 1;
}
int main() {
  int p, m, i;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%s%d", &s[i], &p);
    adj[p].push_back(i);
  }
  Dfs(0, 0);
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d", &v, &k);
    printf("%d\n", solve());
  }
  return 0;
}
