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;
inline int bit(int x, int i) { return (x >> i) & 1; }
int idx[100010];
map<int, int> cnt[100010], que[100010];
vector<int> e[100010];
int fa[17][100010], qv[100010], qp[100010];
int get(int p, int q) {
  for (int i = 16; i >= 0; i--)
    if (bit(q, i)) p = fa[i][p];
  return p;
}
void go(int p, int lv) {
  for (__typeof((e[p]).begin()) it = (e[p]).begin(); it != (e[p]).end(); it++)
    go(*it, lv + 1);
  if (p == 0) return;
  if (e[p].empty())
    idx[p] = p;
  else {
    idx[p] = idx[e[p][0]];
    for (__typeof((e[p]).begin()) it = (e[p]).begin(); it != (e[p]).end(); it++)
      if (((int)(cnt[idx[*it]]).size()) > ((int)(cnt[idx[p]]).size()))
        idx[p] = idx[*it];
  }
  cnt[idx[p]][lv]++;
  for (__typeof((e[p]).begin()) it = (e[p]).begin(); it != (e[p]).end(); it++)
    if (idx[*it] != idx[p])
      for (__typeof((cnt[idx[*it]]).begin()) jt = (cnt[idx[*it]]).begin();
           jt != (cnt[idx[*it]]).end(); jt++)
        cnt[idx[p]][jt->first] += jt->second;
  for (__typeof((que[p]).begin()) it = (que[p]).begin(); it != (que[p]).end();
       it++)
    it->second = cnt[idx[p]][lv + it->first];
}
int main() {
  int n, m, x;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    fa[0][i] = x;
    e[x].push_back(i);
  }
  for (int i = 1; i < 17; i++)
    for (int j = 1; j <= n; j++) fa[i][j] = fa[i - 1][fa[i - 1][j]];
  scanf("%d", &m);
  for (int i = 0; i < m; i++) scanf("%d%d", qv + i, qp + i);
  for (int i = 0; i < m; i++) {
    int p = get(qv[i], qp[i]);
    if (p) que[p][qp[i]] = 0;
  }
  go(0, 0);
  for (int i = 0; i < m; i++) {
    int p = get(qv[i], qp[i]);
    printf("%d%c", p == 0 ? 0 : que[p][qp[i]] - 1, i == m - 1 ? '\n' : ' ');
  }
  return 0;
}