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;
vector<int> g[100010], rt, umr[100010];
int anc[20][100010];
int in[100010], out[100010], dfs_clock, dep[100010];
void dfs(int u, int d) {
  dep[u] = d, in[u] = ++dfs_clock;
  umr[d].push_back(in[u]);
  for (int v : g[u]) dfs(v, d + 1);
  out[u] = dfs_clock;
}
void pre() {
  for (int u : rt) dfs(u, 0);
  for (int j = 1; j <= 17; j++)
    for (int i = 1; i <= n; i++) anc[j][i] = anc[j - 1][anc[j - 1][i]];
}
int cal(int v, int p) {
  int ret = v;
  for (int i = 17; i >= 0; i--)
    if ((1 << i) & p) ret = anc[i][ret];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p)
      g[p].push_back(i), anc[0][i] = p;
    else
      rt.push_back(i);
  }
  pre();
  int q;
  scanf("%d", &q);
  while (q--) {
    int v, p;
    scanf("%d%d", &v, &p);
    v = cal(v, p);
    if (!v) {
      printf("0 ");
      continue;
    }
    int d = dep[v] + p;
    int l = lower_bound(umr[d].begin(), umr[d].end(), in[v]) - umr[d].begin();
    int r = upper_bound(umr[d].begin(), umr[d].end(), out[v]) - umr[d].begin();
    printf("%d ", r - l - 1);
  }
  return 0;
}
