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;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, r[100050], val[100050], ans[100050];
int deep[100050], size[100050], son[100050];
int c[100050];
map<int, int> vis[100050];
struct edge {
  int to, nxt;
} e[100050];
int head[100050], ecnt;
map<string, int> mp;
vector<pair<int, int> > g[100050];
inline void addedge(int u, int v) {
  if (u == 0) return;
  e[ecnt] = (edge){v, head[u]}, head[u] = ecnt++;
}
void DFS3(int u, int key, int ban) {
  vis[deep[u]][val[u]] += key;
  if (key == -1 && vis[deep[u]][val[u]] == 0) c[deep[u]]--;
  if (key == 1 && vis[deep[u]][val[u]] == 1) c[deep[u]]++;
  for (int i = head[u]; ~i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == ban) continue;
    DFS3(v, key, ban);
  }
}
void DFS2(int u, bool flag) {
  for (int i = head[u]; ~i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == son[u]) continue;
    DFS2(v, false);
  }
  if (son[u] != -1) DFS2(son[u], true);
  DFS3(u, 1, son[u]);
  for (int i = 0; i < g[u].size(); i++) ans[g[u][i].second] = c[g[u][i].first];
  if (!flag) DFS3(u, -1, -1);
}
void DFS1(int u, int step) {
  deep[u] = step, size[u] = 1, son[u] = -1;
  for (int i = head[u]; ~i; i = e[i].nxt) {
    int v = e[i].to;
    DFS1(v, step + 1);
    size[u] += size[v];
    if (son[u] == -1 || size[v] > size[son[u]]) son[u] = v;
  }
}
int main() {
  n = read();
  memset(head, -1, sizeof(head));
  for (int i = 1, id = 0; i <= n; i++) {
    char tmp[25];
    scanf("%s", tmp);
    addedge(r[i] = read(), i);
    string s = tmp;
    if (mp[s])
      val[i] = mp[s];
    else
      val[i] = mp[s] = ++id;
  }
  for (int i = 1; i <= n; i++)
    if (r[i] == 0) DFS1(i, 1);
  m = read();
  for (int i = 1; i <= m; i++) {
    int v = read(), k = read() + deep[v];
    if (k <= n) g[v].push_back(make_pair(k, i));
  }
  for (int i = 1; i <= n; i++)
    if (r[i] == 0) DFS2(i, false);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
