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;
int n, m, tot;
int res[100000 + 5], ans[100000 + 5];
int L[100000 + 5], R[100000 + 5], num[100000 + 5];
int dep[100000 + 5], son[100000 + 5], fa[100000 + 5][25];
int top[100000 + 5], siz[100000 + 5];
vector<int> edge[100000 + 5];
struct node {
  int id, deep;
};
vector<node> Q[100000 + 5];
void dfs(int x) {
  L[x] = ++tot;
  num[tot] = x;
  siz[x] = 1, son[x] = 0;
  for (int i = 0; i < (int)edge[x].size(); i++) {
    int y = edge[x][i];
    dep[y] = dep[x] + 1;
    dfs(y);
    siz[x] += siz[y];
    if (siz[y] > siz[son[x]]) son[x] = y;
  }
  R[x] = tot;
}
void Add_point(int x, int d) { res[dep[x]] += d; }
void Add_tree(int x, int d) {
  for (int i = L[x]; i <= R[x]; i++) Add_point(num[i], d);
}
void solve(int x) {
  for (int i = 0; i < (int)edge[x].size(); i++) {
    int y = edge[x][i];
    if (y == son[x]) continue;
    solve(y);
    Add_tree(y, -1);
  }
  if (son[x]) solve(son[x]);
  for (int i = 0; i < (int)edge[x].size(); i++) {
    int y = edge[x][i];
    if (y == son[x]) continue;
    Add_tree(y, 1);
  }
  Add_point(x, 1);
  for (int i = 0; i < (int)Q[x].size(); i++)
    ans[Q[x][i].id] = res[Q[x][i].deep];
}
int Up(int x, int step) {
  for (int i = 0; i <= 25; i++)
    if (step & (1 << i)) x = fa[x][i];
  return x;
}
int main() {
  while (~scanf("%d", &n)) {
    tot = 0;
    for (int i = 0; i <= n; i++) {
      res[i] = 0;
      edge[i].clear();
      Q[i].clear();
    }
    for (int v = 1, u; v <= n; v++) {
      scanf("%d", &u);
      edge[u].push_back(v);
      fa[v][0] = u;
    }
    scanf("%d", &m);
    dep[0] = 0, dfs(0);
    for (int i = 1; i < 22; i++)
      for (int j = 1; j <= n; j++) fa[j][i] = fa[fa[j][i - 1]][i - 1];
    for (int i = 1, u, p; i <= m; i++) {
      scanf("%d%d", &u, &p);
      if (dep[u] - p > 0) {
        int father = Up(u, p);
        Q[father].push_back((node){i, dep[u]});
      } else
        ans[i] = 0;
    }
    solve(0);
    for (int i = 1; i <= m; i++) {
      printf("%d", max(0, ans[i] - 1));
      if (i != m) printf(" ");
    }
  }
}
