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>
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
using namespace std;
int n, m;
struct edge {
  int nxt, to;
} e[101000 << 1];
int head[101000], ecnt;
inline void addedge(int from, int to) {
  e[++ecnt] = (edge){head[from], to};
  head[from] = ecnt;
}
vector<pair<int, int> > vec[101000];
int fa[101000], dep[101000], siz[101000], son[101000];
int f[101000][21];
void dfs_init(int cur) {
  f[cur][0] = fa[cur];
  for (register int j = 1; j <= 19; ++j)
    if (f[cur][j - 1]) {
      f[cur][j] = f[f[cur][j - 1]][j - 1];
    }
  dep[cur] = dep[fa[cur]] + 1;
  siz[cur] = 1;
  int mx = 0;
  for (register int i = head[cur]; i; i = e[i].nxt) {
    int to = e[i].to;
    dfs_init(to);
    if (siz[to] > mx) mx = siz[to], son[cur] = to;
    siz[cur] += siz[to];
  }
}
inline int find(int p, int k) {
  for (register int j = 19; ~j; --j)
    if (k >= (1 << j)) p = f[p][j], k -= (1 << j);
  return p;
}
int bin[101000], ans[101000];
inline void Del(int cur) {
  --bin[dep[cur]];
  for (register int i = head[cur]; i; i = e[i].nxt) {
    int to = e[i].to;
    Del(to);
  }
}
inline void Add(int cur) {
  ++bin[dep[cur]];
  for (register int i = head[cur]; i; i = e[i].nxt) {
    int to = e[i].to;
    Add(to);
  }
}
void Dfs(int cur) {
  ++bin[dep[cur]];
  for (register int i = head[cur]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to == son[cur]) continue;
    Dfs(to);
    Del(to);
  }
  if (son[cur]) Dfs(son[cur]);
  for (register int i = head[cur]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to == son[cur]) continue;
    Add(to);
  }
  for (register unsigned int i = 0; i < vec[cur].size(); ++i) {
    pair<int, int> pr = vec[cur][i];
    ans[pr.second] = bin[pr.first] - 1;
  }
}
int main() {
  read(n);
  for (register int i = 1; i <= n; ++i) {
    read(fa[i]);
    if (fa[i]) addedge(fa[i], i);
  }
  for (register int i = 1; i <= n; ++i)
    if (fa[i] == 0) dfs_init(i);
  read(m);
  for (register int i = 1; i <= m; ++i) {
    int p, k;
    read(p), read(k);
    int anc = find(p, k);
    if (!anc)
      ans[i] = 0;
    else
      vec[anc].push_back(make_pair(dep[p], i));
  }
  for (register int i = 1; i <= n; ++i)
    if (fa[i] == 0) Dfs(i), Del(i);
  for (register int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}