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;
const int maxn = 3e5 + 1;
int sons[maxn], res[maxn], L[maxn];
bool heavy[maxn];
string s[maxn];
int n, m, l;
map<string, int> mp[maxn];
vector<int> ady[maxn];
vector<pair<int, int> > query[maxn];
void dfs(int u, int p) {
  sons[u] = 1;
  L[u] = L[p] + 1;
  if (L[u] > l) l = L[u];
  for (auto v : ady[u]) {
    if (v != p) {
      dfs(v, u);
      sons[u] += sons[v];
    }
  }
}
void add(int u, int p, int level, int x) {
  mp[level][s[u]] += x;
  if (mp[level][s[u]] == 0) mp[level].erase(s[u]);
  for (auto v : ady[u])
    if (v != p && !heavy[v]) add(v, u, level + 1, x);
}
void dsu(int u, int p, bool mark) {
  int len = -1, big = -1;
  for (auto v : ady[u])
    if (v != p && sons[v] > len) len = sons[v], big = v;
  for (auto v : ady[u])
    if (v != p && v != big) dsu(v, u, false);
  if (big != -1) dsu(big, u, true), heavy[big] = true;
  add(u, p, L[u], 1);
  int c;
  for (auto i : query[u]) {
    res[i.second] = mp[L[u] + i.first].size();
  }
  if (big != -1) heavy[big] = false;
  if (!mark) add(u, p, L[u], -1);
}
int main() {
  scanf("%d", &n);
  int a, b;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    scanf("%d", &a);
    ady[a].push_back(i);
    ady[i].push_back(a);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &a, &b);
    query[a].push_back({b, i});
  }
  dfs(0, 0);
  dsu(0, 0, 0);
  for (int i = 1; i <= m; i++) printf("%d\n", res[i]);
}
