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 maxn = 1e5 + 10;
struct E {
  int to;
  int next;
} ed[maxn];
int head[maxn];
int tot;
struct D {
  char s[30];
} a[maxn];
void J(int a, int b) {
  ed[++tot].to = b;
  ed[tot].next = head[a];
  head[a] = tot;
}
int deep[maxn];
int son[maxn];
int vis[maxn];
int ans[maxn];
int size[maxn];
void Dfs(int x, int fa) {
  deep[x] = deep[fa] + 1;
  size[x] = 1;
  for (int i = head[x]; i; i = ed[i].next) {
    Dfs(ed[i].to, x);
    size[x] += size[ed[i].to];
    if (!son[x] || size[ed[i].to] > size[son[x]]) son[x] = ed[i].to;
  }
}
int nrt[maxn];
struct Node {
  int k;
  int pst;
  Node(int a, int b) {
    k = a;
    pst = b;
  }
};
vector<Node> ve[maxn];
struct SS {
  char s[30];
  int deep;
  SS(char a[], int b) {
    for (int i = 0; i < 30; i++) s[i] = a[i];
    deep = b;
  }
  friend bool operator<(SS a, SS b) {
    if (a.deep == b.deep) {
      for (int i = 0; i <= 19; i++)
        if (a.s[i] != b.s[i]) return a.s[i] < b.s[i];
      return 0;
    }
    return a.deep < b.deep;
  }
};
map<SS, int> mapp;
int tmp[maxn];
void up(int x) {
  if (!mapp[SS(a[x].s, deep[x])]) {
    mapp[SS(a[x].s, deep[x])] = 1;
    tmp[deep[x]]++;
  }
  for (int i = head[x]; i; i = ed[i].next) up(ed[i].to);
}
void get_ans(int x) {
  for (int i = 0; i < ve[x].size(); i++)
    ans[ve[x][i].pst] = tmp[deep[x] + ve[x][i].k];
}
void C(int x) {
  tmp[deep[x]] = 0;
  for (int i = head[x]; i; i = ed[i].next) C(ed[i].to);
}
void Dfs_(int x, int t) {
  for (int i = head[x]; i; i = ed[i].next) {
    if (ed[i].to == son[x]) continue;
    Dfs_(ed[i].to, 0);
  }
  if (son[x]) Dfs_(son[x], 1);
  if (!mapp[SS(a[x].s, deep[x])]) {
    mapp[SS(a[x].s, deep[x])] = 1;
    tmp[deep[x]]++;
  }
  for (int i = head[x]; i; i = ed[i].next) {
    if (ed[i].to == son[x]) continue;
    up(ed[i].to);
  }
  get_ans(x);
  if (!t) {
    mapp.clear();
    C(x);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int x;
  for (int i = 1; i <= n; i++) {
    scanf("%s%d", a[i].s, &x);
    nrt[i] = x;
    if (x) J(x, i);
  }
  for (int i = 1; i <= n; i++)
    if (!size[i]) Dfs(i, 0);
  int m;
  scanf("%d", &m);
  int pst;
  int k;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &pst, &k);
    if (deep[pst] + k > n) continue;
    ve[pst].push_back(Node(k, i));
  }
  for (int i = 1; i <= n; i++)
    if (!nrt[i]) Dfs_(i, 0);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
