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 P = 131;
struct Query {
  int id, k;
};
int n, m, a[100010], cnt, ans[100010];
int d[100010], sz[100010], son[100010], sum[100010];
int hson;
vector<Query> q[100010];
vector<int> v[100010];
unordered_map<unsigned long long, int> mp;
unordered_map<int, int> vis[100010];
void read(int& x) {
  char c = getchar();
  x = 0;
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
}
void dfs(int x) {
  sz[x] = 1;
  for (int y : v[x]) {
    d[y] = d[x] + 1;
    dfs(y);
    if (sz[son[x]] < sz[y]) son[x] = y;
    sz[x] += sz[y];
  }
}
inline void add(int x, int op) {
  if (!vis[d[x]][a[x]]) sum[d[x]] += op;
  vis[d[x]][a[x]] += op;
  if (!vis[d[x]][a[x]]) sum[d[x]] += op;
}
void update(int x, int op) {
  add(x, op);
  for (int y : v[x])
    if (y != hson) update(y, op);
}
void dfs(int x, bool keep) {
  for (int y : v[x])
    if (y != son[x]) dfs(y, 0);
  if (son[x]) dfs(son[x], 1);
  hson = son[x];
  update(x, 1);
  hson = 0;
  for (Query qry : q[x]) ans[qry.id] = sum[d[x] + qry.k];
  if (!keep) update(x, -1);
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    char s[25];
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    unsigned long long hash = 0;
    for (int j = 1; j <= len; j++) hash = hash * P + s[j];
    if (!mp[hash])
      mp[hash] = a[i] = ++cnt;
    else
      a[i] = mp[hash];
    int x;
    read(x);
    v[x].push_back(i);
  }
  dfs(0);
  read(m);
  for (int i = 1; i <= m; i++) {
    int x, k;
    read(x);
    read(k);
    if (d[x] + k <= n) q[x].push_back((Query){i, k});
  }
  dfs(0, 0);
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
}