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;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline string get() {
  char ch[100];
  scanf("%s", ch);
  return ch;
}
const int MAX_LG = 21;
const int maxn = 5e5 + 10;
const int mod = 1e9 + 7;
int root;
vector<int> g[maxn];
vector<pair<int, int> > g2[maxn];
vector<pair<string, int> > vc;
int cmp[maxn], sz[maxn], lvl[maxn], res[maxn];
bool mark[maxn];
map<int, int> cnt[maxn];
int cnt2[maxn];
inline void calc(int v, int pr = -1) {
  sz[v] = 1;
  for (auto u : g[v])
    if (u - pr) lvl[u] = lvl[v] + 1, calc(u, v), sz[v] += sz[u];
}
inline void add(int v, int pr, int val) {
  if ((cnt[lvl[v]][cmp[v]] == 1 && val == -1) ||
      (cnt[lvl[v]][cmp[v]] == 0 && val == 1))
    cnt2[lvl[v]] += val;
  cnt[lvl[v]][cmp[v]] += val;
  for (auto u : g[v])
    if (u - pr && !mark[u]) add(u, v, val);
}
inline void dfs(int v, int pr = -1, int del = 1) {
  int maxi = -1, id = -1;
  for (auto u : g[v])
    if (u - pr && sz[u] > maxi) maxi = sz[u], id = u;
  for (auto u : g[v])
    if (u - pr && u - id) dfs(u, v);
  if (id + 1) dfs(id, v, 0), mark[id] = true;
  add(v, pr, 1);
  for (int i = 0; i < g2[v].size(); i++)
    res[g2[v][i].second] = cnt2[g2[v][i].first];
  if (id + 1) mark[id] = false;
  if (del) add(v, pr, -1);
}
int32_t main() {
  int n = in();
  root = 0;
  for (int i = 0; i < n; i++) {
    string name = get();
    int parent = in();
    g[parent].push_back(i + 1), g[i + 1].push_back(parent);
    vc.push_back({name, i + 1});
  }
  sort(vc.begin(), vc.end());
  for (int i = 0; i < vc.size(); i++)
    cmp[vc[i].second] =
        (i && vc[i].first == vc[i - 1].first ? cmp[vc[i - 1].second] : i);
  int m = in();
  calc(root);
  for (int i = 0; i < m; i++) {
    int v = in(), h = lvl[v] + in();
    g2[v].push_back({h, i});
  }
  dfs(root);
  for (int i = 0; i < m; i++) printf("%d\n", res[i]);
}
