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;
vector<int> v[100005];
int n;
int parent[100005];
int in[100005];
int depth[100005];
int cnt[100005];
int mark;
vector<int> h[100005];
vector<int> id[100005];
void dfs(int x, int hg, int par) {
  int sz = v[x].size();
  int i, j;
  h[hg].push_back(mark);
  id[hg].push_back(x);
  in[x] = mark;
  depth[x] = hg;
  cnt[x] = 1;
  mark++;
  for (i = 0; i < sz; i++) {
    int y = v[x][i];
    if (y == par) continue;
    dfs(y, hg + 1, x);
    cnt[x] += cnt[y];
  }
}
int bs1(int indx, int val) {
  int ret = -1, l = 0, r = h[indx].size() - 1;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (h[indx][mid] <= val) {
      ret = mid;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  return ret + 1;
}
int main() {
  scanf("%d", &n);
  int i, j;
  int x;
  for (i = 1; i <= n; i++) {
    scanf("%d", &parent[i]);
    if (parent[i] == 0) continue;
    x = parent[i];
    v[i].push_back(x);
    v[x].push_back(i);
  }
  mark = 0;
  for (i = 1; i <= n; i++) {
    if (parent[i] == 0) {
      dfs(i, 0, -1);
    }
  }
  int m;
  scanf("%d", &m);
  int y;
  for (i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    if (depth[x] < y) {
      printf("0 ");
    } else {
      int inh = depth[x] - y;
      int par = bs1(inh, in[x]);
      par = id[inh][par - 1];
      int cnt1 = bs1(depth[x], in[par] + cnt[par] - 1);
      int cnt2 = bs1(depth[x], in[par] - 1);
      int ans = cnt1 - cnt2 - 1;
      printf("%d ", ans);
    }
  }
  printf("\n");
  return 0;
}
