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>
const int mod = 998244353;
const int maxn = 1e5 + 5;
const int inf = 1e9;
const long long onf = 1e18;
using namespace std;
int n, m;
int bigson;
int Bigson[maxn], Size[maxn], sum[maxn << 1];
int father[maxn], depth[maxn];
int ans[maxn];
string str[maxn];
map<string, int> a[maxn];
struct Node {
  int i, dep;
};
vector<Node> q[maxn];
vector<int> maps[maxn];
void dfs1(int u) {
  Size[u] = 1;
  for (auto v : maps[u]) {
    depth[v] = depth[u] + 1;
    dfs1(v);
    Size[u] += Size[v];
    if (Size[v] > Size[Bigson[u]]) Bigson[u] = v;
  }
}
void insert(int u, int val) {
  if (val) {
    a[depth[u]][str[u]]++;
    if (a[depth[u]][str[u]] == 1) sum[depth[u]]++;
  } else {
    a[depth[u]][str[u]]--;
    if (a[depth[u]][str[u]] == 0) sum[depth[u]]--;
  }
  for (auto v : maps[u]) {
    if (v == bigson) continue;
    insert(v, val);
  }
}
void dfs2(int u, int opt) {
  for (auto v : maps[u]) {
    if (v == Bigson[u]) continue;
    dfs2(v, 0);
  }
  if (Bigson[u]) dfs2(Bigson[u], 1), bigson = Bigson[u];
  insert(u, 1);
  bigson = -1;
  for (auto it : q[u]) ans[it.i] = sum[it.dep + depth[u]];
  if (!opt) insert(u, 0);
}
void work() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> str[i] >> father[i];
    maps[father[i]].push_back(i);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int v, h;
    scanf("%d%d", &v, &h);
    q[v].push_back(Node{i, h});
  }
  for (int i = 1; i <= n; i++) {
    if (father[i] == 0) dfs1(i);
  }
  for (int i = 1; i <= n; i++) {
    if (father[i] == 0) dfs2(i, 0);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
int main() {
  int t = 1, Case = 1;
  while (t--) {
    work();
  }
  return 0;
}
