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>
int const maxn = 1e5 + 10;
int const maxm = 1e5 + 10;
using namespace std;
unordered_map<int, int> cnt[maxn];
unordered_map<string, int> p;
int id[maxn];
int idx = 0;
vector<pair<int, int> > query[maxn];
int ans[maxn];
int head[maxn], tot;
struct Edg {
  int to, next;
} edg[maxm * 2];
void inint() {
  memset(head, -1, sizeof head);
  tot = 0;
}
void add(int u, int v) {
  edg[tot].next = head[u];
  edg[tot].to = v;
  head[u] = tot++;
}
int dep[maxn], son[maxn], sz[maxn];
void dfs1(int x, int fa) {
  sz[x] = 1;
  int maxval = 0;
  for (int i = head[x]; ~i; i = edg[i].next) {
    int y = edg[i].to;
    dep[y] = dep[x] + 1;
    dfs1(y, x);
    sz[x] += sz[y];
    if (sz[y] > maxval) {
      maxval = sz[y];
      son[x] = y;
    }
  }
}
void erase(int x, int fa) {
  cnt[dep[x]][id[x]]--;
  if (cnt[dep[x]][id[x]] == 0) cnt[dep[x]].erase(id[x]);
  for (int i = head[x]; ~i; i = edg[i].next) {
    int y = edg[i].to;
    erase(y, x);
  }
}
void get(int x, int fa, int p) {
  cnt[dep[x]][id[x]]++;
  for (int i = head[x]; ~i; i = edg[i].next) {
    int y = edg[i].to;
    if (y == p) continue;
    get(y, x, p);
  }
}
void dfs(int x, int fa) {
  for (int i = head[x]; ~i; i = edg[i].next) {
    int y = edg[i].to;
    if (y == son[x]) continue;
    dfs(y, x);
    erase(y, x);
  }
  if (son[x]) dfs(son[x], x);
  get(x, fa, son[x]);
  for (auto now : query[x]) {
    ans[now.first] = cnt[now.second].size();
  }
}
int main() {
  int n;
  scanf("%d", &n);
  string ch;
  int pre;
  inint();
  for (int i = 1; i <= n; i++) {
    cin >> ch;
    if (p.count(ch) == 0) {
      p[ch] = ++idx;
    }
    id[i] = p[ch];
    scanf("%d", &pre);
    add(pre, i);
  }
  int m;
  scanf("%d", &m);
  dfs1(0, -1);
  for (int i = 1; i <= m; i++) {
    int v, k;
    scanf("%d%d", &v, &k);
    if (dep[v] + k > 1e5) {
      ans[i] = 0;
      continue;
    }
    query[v].push_back({i, dep[v] + k});
  }
  dfs(0, -1);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
