Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f == 1 ? x : -x;
}
const int N = 2e5 + 4;
int n, Q, tot, siz[N], son[N], dep[N], ans[N], vis[N], a[N], fa[N];
vector<int> e[N];
vector<pair<int, int> > q[N];
map<int, bool> mp[N];
map<string, int> mp2;
void dfs_1(int x, int fa) {
  siz[x] = 1;
  dep[x] = dep[fa] + 1;
  for (auto v : e[x]) {
    if (v == fa) continue;
    dfs_1(v, x);
    siz[x] += siz[v];
    if (siz[v] > siz[son[x]]) son[x] = v;
  }
}
void calc(int x, int fa, int fl) {
  if (fl == 1)
    mp[dep[x]][a[x]] = 1;
  else
    mp[dep[x]].erase(a[x]);
  for (auto v : e[x])
    if (v != fa && !vis[v]) calc(v, x, fl);
}
void dfs_2(int x, int fa, int keep) {
  for (auto v : e[x])
    if (v != fa && v != son[x]) dfs_2(v, x, 0);
  if (son[x]) {
    dfs_2(son[x], x, 1);
    vis[son[x]] = 1;
  }
  calc(x, fa, 1);
  for (auto v : q[x]) {
    ans[v.second] = mp[v.first].size();
  }
  if (son[x]) vis[son[x]] = 0;
  if (!keep) calc(x, fa, -1);
}
string s;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    cin >> s;
    if (!mp2[s])
      a[i] = mp2[s] = ++tot;
    else
      a[i] = mp2[s];
    e[fa[i] = read()].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    if (!fa[i]) dfs_1(i, 0);
  Q = read();
  for (int i = 1, x; i <= Q; i++) {
    x = read();
    q[x].push_back(make_pair(dep[x] + read(), i));
  }
  for (int i = 1; i <= n; i++)
    if (!fa[i]) dfs_2(i, 0, 0);
  for (int i = 1; i <= Q; i++) cout << ans[i] << "\n";
  return (0 - 0);
}