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;
const int N = 500005;
const int LN = 31;
int ancestor[N][LN], vis[N], en[N], strt[N], lvl[N];
int time1;
vector<int> adj[N], dep[N];
void dfs(int st, int cnt) {
  vis[st] = 1;
  strt[st] = ++time1;
  lvl[st] = cnt;
  dep[lvl[st]].push_back(time1);
  for (auto i : adj[st]) {
    if (!vis[i]) dfs(i, cnt + 1);
  }
  en[st] = ++time1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ancestor[i][0]);
    adj[ancestor[i][0]].push_back(i);
  }
  for (int j = 1; j < LN; j++) {
    for (int i = 1; i <= n; i++) {
      ancestor[i][j] = ancestor[ancestor[i][j - 1]][j - 1];
    }
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 1);
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b, cnt = 0;
    scanf("%d", &a), scanf("%d", &b);
    cnt = a;
    for (int i = 0; i < LN; i++) {
      if (b & (1 << i)) a = ancestor[a][i];
    }
    if (a == 0) {
      cout << 0 << endl;
      continue;
    }
    int idx = upper_bound(dep[lvl[cnt]].begin(), dep[lvl[cnt]].end(), en[a]) -
              lower_bound(dep[lvl[cnt]].begin(), dep[lvl[cnt]].end(), strt[a]);
    printf("%d\n", idx - 1);
  }
  return 0;
}
