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;
int n, m;
vector<int> to[100005];
int deep[100005];
int size[100005];
int son[100005];
int nid[100005];
int id[100005];
int cnt;
string name[100005];
vector<pair<int, int> > que[100005];
int type[200005];
map<string, int> mp[100005];
int ans[100005];
void updata(int x, int k) {
  if (mp[deep[x]][name[x]] == 0) type[deep[x]]++;
  mp[deep[x]][name[x]] += k;
  if (mp[deep[x]][name[x]] == 0) type[deep[x]]--;
}
void dfs1(int now) {
  size[now] = 1;
  id[now] = ++cnt;
  nid[cnt] = now;
  int mx = 0;
  for (auto next : to[now]) {
    deep[next] = deep[now] + 1;
    dfs1(next);
    size[now] += size[next];
    if (size[next] > mx) {
      mx = size[next];
      son[now] = next;
    }
  }
}
void dfs2(int now, int keep) {
  for (auto next : to[now]) {
    if (next == son[now]) continue;
    dfs2(next, 0);
  }
  if (son[now]) dfs2(son[now], 1);
  for (auto next : to[now]) {
    if (next == son[now]) continue;
    for (int i = 0; i < size[next]; i++) {
      updata(nid[id[next] + i], 1);
    }
  }
  updata(now, 1);
  for (auto Q : que[now]) {
    int dp = Q.first;
    int di = Q.second;
    ans[di] = type[deep[now] + dp];
  }
  if (keep == 0) {
    for (int i = 0; i < size[now]; i++) {
      updata(nid[id[now] + i], -1);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> name[i];
    int p;
    scanf("%d", &p);
    to[p].push_back(i);
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    que[a].push_back(make_pair(b, i));
  }
  dfs1(0);
  dfs2(0, 0);
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}