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;
template <class T>
inline void rd(T &x) {
  x = 0;
  char c;
  int f = 1;
  while (c = getchar(), c < 48)
    if (c == '-') f = -1;
  do x = x * 10 + (c & 15);
  while (c = getchar(), c > 47);
  x *= f;
}
template <class T>
void pf(T x) {
  if (x < 0)
    putchar('-'), x = -x;
  else if (!x) {
    putchar('0');
    return;
  }
  static int stk[100], tp;
  for (tp = 0; x; x /= 10) stk[tp++] = x % 10;
  while (tp) putchar(stk[--tp] ^ 48);
}
template <class T>
void ptn(T x) {
  pf(x), putchar('\n');
}
template <class T>
void ptk(T x) {
  pf(x), putchar(' ');
}
template <class T>
void Min(T &A, T B) {
  if (A > B) A = B;
}
template <class T>
void Max(T &A, T B) {
  if (A < B) A = B;
}
const int N = (int)1e5 + 5;
map<int, int> mp[N];
map<int, int>::iterator it;
int head[N], val[N];
pair<int, int> G[N];
inline void add_edge(int x, int y) {
  static int tot_edge;
  G[tot_edge] = pair<int, int>(y, head[x]);
  head[x] = tot_edge++;
}
vector<pair<int, int> > Q[N];
map<string, int> ID;
int n, m, ID_cnt, Ans[N], sz[N], son[N], Lpos[N], Rpos[N], Mp[N], dep[N];
void dfs_pre(int x) {
  static int dfs_clock;
  sz[Mp[Lpos[x] = ++dfs_clock] = x] = 1;
  for (int i = head[x]; ~i; i = G[i].second) {
    dep[G[i].first] = dep[x] + 1;
    dfs_pre(G[i].first);
    if (sz[son[x]] < sz[G[i].first]) son[x] = G[i].first;
    sz[x] += sz[G[i].first];
  }
  Rpos[x] = dfs_clock + 1;
}
void add(int x) {
  for (int i = (Lpos[x]), _t = (Rpos[x]); i < _t; ++i)
    mp[dep[Mp[i]]][val[Mp[i]]]++;
}
void del(int x) {
  for (int i = (Lpos[x]), _t = (Rpos[x]); i < _t; ++i) {
    it = mp[dep[Mp[i]]].find(val[Mp[i]]);
    assert(it->second);
    if (!(--it->second)) mp[dep[Mp[i]]].erase(it);
  }
}
void solve(int x) {
  for (int i = head[x]; ~i; i = G[i].second)
    if (son[x] != G[i].first) solve(G[i].first), del(G[i].first);
  if (son[x]) solve(son[x]);
  for (int i = head[x]; ~i; i = G[i].second)
    if (son[x] != G[i].first) add(G[i].first);
  mp[dep[x]][val[x]]++;
  for (int i = (0), _t = (Q[x].size()); i < _t; ++i)
    Ans[Q[x][i].second] = mp[Q[x][i].first].size();
}
int main() {
  memset(head, -1, sizeof head);
  rd(n);
  string a;
  for (int i = 1, fa; i <= n; ++i) {
    cin >> a;
    rd(fa);
    add_edge(fa, i);
    if (!ID[a]) ID[a] = ++ID_cnt;
    val[i] = ID[a];
  }
  dfs_pre(0);
  rd(m);
  for (int i = 1, x, y; i <= m; ++i) {
    rd(x);
    rd(y);
    y += dep[x];
    if (y <= n) Q[x].push_back(pair<int, int>(y, i));
  }
  solve(0);
  for (int i = (1), _t = (m + 1); i < _t; ++i) ptn(Ans[i]);
  return 0;
}