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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f = ch ^ 45, ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return f ? x : -x;
}
int ls[100010 * 18], rs[100010 * 18], rt[100010];
int dep[100010], frt[100010], nxt[100010], v[100010];
set<int> t[100010 * 18];
map<string, int> ma;
int s[100010];
int n, m, maxd, cnt, tot, c;
int ans[100010], q[100010];
vector<int> vc[100010], ve[100010];
inline void add(int x, int y) {
  v[++tot] = y;
  nxt[tot] = frt[x];
  frt[x] = tot;
}
void getdep(int x) {
  maxd = max(maxd, dep[x]);
  for (int i = frt[x]; i; i = nxt[i]) {
    dep[v[i]] = dep[x] + 1;
    getdep(v[i]);
  }
}
inline int work(int x, int y) {
  if (t[x].size() < t[y].size()) x ^= y ^= x ^= y;
  for (set<int>::iterator it = t[y].begin(); it != t[y].end(); it++)
    t[x].insert(*it);
  return x;
}
void ins(int &now, int l, int r, int x, int k) {
  if (!now) now = ++cnt;
  if (l == r) {
    t[now].insert(s[k]);
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    ins(ls[now], l, mid, x, k);
  else
    ins(rs[now], mid + 1, r, x, k);
}
int merge(int x, int y, int l, int r) {
  if (!x || !y) return x ^ y;
  if (l == r) return work(x, y);
  int mid = l + r >> 1;
  ls[x] = merge(ls[x], ls[y], l, mid);
  rs[x] = merge(rs[x], rs[y], mid + 1, r);
  return x;
}
int query(int l, int r, int x, int now) {
  if (x > r) return 0;
  if (!now) return 0;
  if (l == r) {
    return t[now].size();
  }
  int mid = l + r >> 1;
  if (x <= mid)
    return query(l, mid, x, ls[now]);
  else
    return query(mid + 1, r, x, rs[now]);
}
void dfs(int x) {
  for (int i = frt[x]; i; i = nxt[i]) {
    dfs(v[i]);
    rt[x] = merge(rt[x], rt[v[i]], 1, maxd);
  }
  ins(rt[x], 1, maxd, dep[x], x);
  for (int i = 0; i < vc[x].size(); ++i)
    ans[ve[x][i]] = query(1, maxd, dep[x] + vc[x][i], rt[x]);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    string str;
    cin >> str;
    if (!ma[str]) ma[str] = ++c;
    s[i] = ma[str];
    int x = read();
    add(x, i);
  }
  getdep(0);
  m = read();
  for (int i = 1; i <= m; ++i) {
    int x = read();
    ve[x].push_back(i);
    vc[x].push_back(read());
  }
  dfs(0);
  for (int i = 1; i <= m; ++i) cout << ans[i] << '\n';
}
