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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < 48 || c > 57; c = getchar())
    if (!(c ^ 45)) f = 1;
  for (; c >= 48 && c <= 57; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  f ? x = -x : 0;
}
struct edge {
  int to, nxt;
} e[200005];
int n, m, et, head[100005], a[100005], sz[100005], dep[100005], sn[100005],
    gx[200005], rs[100005];
map<int, int> v[100005];
map<string, int> wq;
char avd[100005];
vector<pair<int, int> > q[100005];
inline void adde(int x, int y) { e[++et] = (edge){y, head[x]}, head[x] = et; }
inline void dfs1(int x, int fa) {
  dep[x] = dep[fa] + 1, sz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    dfs1(e[i].to, x), sz[x] += sz[e[i].to];
    if (sz[sn[x]] < sz[e[i].to]) sn[x] = e[i].to;
  }
}
inline void baoli(int x, int tag) {
  if (tag) {
    v[dep[x]][a[x]]++;
    if (v[dep[x]][a[x]] == 1) gx[dep[x]]++;
  } else {
    v[dep[x]][a[x]]--;
    if (!v[dep[x]][a[x]]) gx[dep[x]]--;
  }
  for (int i = head[x]; i; i = e[i].nxt)
    if (!avd[e[i].to]) baoli(e[i].to, tag);
}
inline void solve(int x, int isw) {
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != sn[x]) solve(e[i].to, 0);
  {
    if (sn[x]) solve(sn[x], 1), avd[sn[x]] = 1;
  }
  baoli(x, 1), avd[sn[x]] = 0;
  for (int i = 0; i < (int)q[x].size(); i++)
    rs[q[x][i].second] = gx[q[x][i].first + dep[x]];
  if (!isw) baoli(x, 0);
}
int main() {
  read(n);
  string s;
  int Id = 0, rt = n + 1;
  for (int i = 1, f; i <= n; i++) {
    cin >> s, read(f), f ? adde(f, i) : adde(rt, i);
    if (wq[s])
      a[i] = wq[s];
    else
      a[i] = wq[s] = ++Id;
  }
  read(m);
  for (int i = 1, x, y; i <= m; i++)
    read(x), read(y), q[x].push_back(make_pair(y, i));
  dfs1(rt, 0), solve(rt, 1);
  for (int i = 1; i <= m; i++) printf("%d\n", rs[i]);
  return 0;
}
