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;
const int N = 1e5 + 10;
string s;
int n, m, E, tot;
int fa[N], fir[N], nex[N << 1], arr[N << 1];
int dep[N], siz[N], son[N], name[N], ans[N];
set<int> Set[N];
map<int, int> Map[N];
vector<pair<int, int> > vec[N];
map<string, int> hs;
inline void Add_Edge(int x, int y) {
  nex[++E] = fir[x];
  fir[x] = E;
  arr[E] = y;
}
void dfs(int x) {
  siz[x] = 1;
  for (int i = fir[x]; i; i = nex[i]) {
    dep[arr[i]] = dep[x] + 1;
    dfs(arr[i]);
    siz[x] += siz[arr[i]];
    if (siz[arr[i]] > siz[son[x]]) son[x] = arr[i];
  }
}
inline void Add(int x, int d) {
  if (++Map[d][x] == 1) Set[d].insert(x);
}
inline void Del(int x, int d) {
  if (--Map[d][x] == 0) Set[d].erase(x);
}
void Updata(int x, int type) {
  if (type == 1)
    Add(name[x], dep[x]);
  else
    Del(name[x], dep[x]);
  for (int i = fir[x]; i; i = nex[i]) Updata(arr[i], type);
}
void Solve(int x, int keep) {
  for (int i = fir[x]; i; i = nex[i]) {
    if (arr[i] == son[x]) continue;
    Solve(arr[i], 0);
  }
  if (son[x]) Solve(son[x], 1);
  for (int i = fir[x]; i; i = nex[i]) {
    if (arr[i] == son[x]) continue;
    Updata(arr[i], 1);
  }
  Add(name[x], dep[x]);
  for (int i = 0; i < vec[x].size(); i++) {
    ans[vec[x][i].first] = Set[vec[x][i].second].size();
  }
  if (!keep) {
    Del(name[x], dep[x]);
    for (int i = fir[x]; i; i = nex[i]) Updata(arr[i], -1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> s >> fa[i];
    if (!hs[s]) hs[s] = ++tot;
    name[i] = hs[s];
    Add_Edge(fa[i], i);
  }
  scanf("%d", &m);
  for (int i = 1; i <= n; i++)
    if (!fa[i]) dep[i] = 1, dfs(i);
  for (int i = 1; i <= m; i++) {
    int x, d;
    scanf("%d%d", &x, &d);
    vec[x].push_back(make_pair(i, dep[x] + d));
  }
  for (int i = 1; i <= n; i++)
    if (!fa[i]) Solve(i, 0);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
