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 sum = 0, p = 1;
  char ch = getchar();
  while (!(('0' <= ch && ch <= '9') || ch == '-')) ch = getchar();
  if (ch == '-') p = -1, ch = getchar();
  while ('0' <= ch && ch <= '9') sum = sum * 10 + ch - 48, ch = getchar();
  return sum * p;
}
const int maxn = 1e5 + 20;
struct node {
  int v, next;
};
node e[maxn];
int cnt, start[maxn];
inline void addedge(int u, int v) {
  e[++cnt] = (node){v, start[u]};
  start[u] = cnt;
}
int n, m;
int p[maxn][20];
struct Query {
  int x, dep;
};
Query qu[maxn];
vector<int> to[maxn];
int rt[maxn], tot;
inline void init() {
  n = read();
  for (int i = (1), _end_ = (n); i <= _end_; i++) {
    int &x = p[i][0];
    x = read();
    if (x)
      addedge(x, i);
    else
      rt[++tot] = i;
  }
  for (int j = (1), _end_ = (19); j <= _end_; j++)
    for (int i = (1), _end_ = (n); i <= _end_; i++)
      p[i][j] = p[p[i][j - 1]][j - 1];
  m = read();
  for (int i = (1), _end_ = (m); i <= _end_; i++) {
    int x = read(), dep = read();
    for (int j = (19), _end_ = (0); j >= _end_; j--)
      if (dep >> j & 1) x = p[x][j];
    qu[i] = (Query){x, dep};
    to[x].push_back(i);
  }
}
int deep[maxn], sz[maxn], son[maxn];
int s[maxn];
void dfs(int u) {
  sz[u] = 1;
  for (int i = start[(u)]; i; i = e[i].next) {
    int v = e[i].v;
    deep[v] = deep[u] + 1;
    dfs(v);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}
int ans[maxn];
inline void Add(int x, int d) { s[deep[x]] += d; }
void update(int u, int fa, int op) {
  Add(u, op);
  for (int i = start[(u)]; i; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    update(v, u, op);
  }
}
void dfs1(int u, int fa) {
  for (int i = start[(u)]; i; i = e[i].next) {
    int v = e[i].v;
    if (son[u] == v) continue;
    dfs1(v, u);
  }
  if (son[u]) dfs1(son[u], u);
  for (int i = start[(u)]; i; i = e[i].next) {
    int v = e[i].v;
    if (v == son[u]) continue;
    update(v, u, 1);
  }
  Add(u, 1);
  for (int v : to[u]) {
    int dep = deep[u] + qu[v].dep;
    ans[v] = s[dep] - 1;
  }
  if (son[fa] != u) update(u, fa, -1);
}
inline void doing() {
  deep[1] = 1;
  for (int i = (1), _end_ = (tot); i <= _end_; i++) dfs(rt[i]);
  for (int i = (1), _end_ = (tot); i <= _end_; i++) dfs1(rt[i], 0);
  for (int i = (1), _end_ = (m); i <= _end_; i++) printf("%d ", ans[i]);
}
int main() {
  init();
  doing();
  return 0;
}
