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;
const int N = 100005;
int fa[N][20], dep[N], num[N], en[N], cnt;
vector<int> a[N], e[N];
void dfs(int u) {
  for (int i = 1; (1 << i) < dep[u]; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  num[u] = ++cnt;
  a[dep[u]].push_back(num[u]);
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    dep[v] = dep[u] + 1;
    dfs(v);
  }
  en[u] = cnt;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &fa[i][0]);
    e[fa[i][0]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (!fa[i][0]) {
      dep[i] = 1;
      dfs(i);
    }
  }
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int v, p;
    scanf("%d%d", &v, &p);
    int x = v;
    for (int j = 17; j >= 0; j--)
      if (dep[fa[x][j]] >= dep[v] - p) x = fa[x][j];
    int ans = upper_bound(a[dep[v]].begin(), a[dep[v]].end(), en[x]) -
              lower_bound(a[dep[v]].begin(), a[dep[v]].end(), num[x]) - 1;
    printf("%d ", max(ans, 0));
  }
  return 0;
}