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 N = 100009;
int n, m, son[N], p[N], sz[N], d[N], ltr[N], cUnq[N], ans[N], iName[N];
set<int> S[N * 2];
int timer, tI[N], tO[N], id[N];
int hd[N * 2], to[N * 2], nxt[N * 2], nE;
struct query {
  int id, k;
};
vector<query> q[N];
void addedge(int u, int v) { to[++nE] = v, nxt[nE] = hd[u], hd[u] = nE; }
void dfs_pre(int u) {
  d[u] = d[p[u]] + 1;
  sz[u] = 1;
  son[u] = 0;
  id[tI[u] = ++timer] = u;
  for (int i = hd[u]; i; i = nxt[i]) {
    int v = to[i];
    dfs_pre(v);
    sz[u] += sz[v];
    if (sz[son[u]] < sz[v]) son[u] = v;
  }
  tO[u] = timer;
}
void aNode(int u, bool tag) {
  if (tag)
    S[d[u]].insert(iName[u]);
  else
    S[d[u]].erase(iName[u]);
}
void aTree(int u, bool tag) {
  for (int i = tI[u]; i <= tO[u]; ++i) aNode(id[i], tag);
}
void dfs(int u, bool hvy) {
  for (int i = hd[u]; i; i = nxt[i])
    if (son[u] != to[i]) dfs(to[i], 0);
  if (son[u]) dfs(son[u], 1);
  for (int i = hd[u]; i; i = nxt[i])
    if (son[u] != to[i]) aTree(to[i], 1);
  aNode(u, 1);
  for (int i = 0; i < q[u].size(); ++i)
    ans[q[u][i].id] = S[d[u] + q[u][i].k].size();
  if (hvy) return;
  aTree(u, 0);
}
void input() {
  map<string, int> names;
  int nNames = 0;
  scanf("%d", &n);
  for (int u = 1; u <= n; ++u) {
    string s;
    cin >> s;
    if (names.count(s))
      iName[u] = names[s];
    else
      iName[u] = names[s] = ++nNames;
    scanf("%d", &p[u]);
    if (p[u]) addedge(p[u], u);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int v, k;
    scanf("%d %d", &v, &k);
    q[v].push_back((query){i, k});
  }
}
void solve() {
  for (int u = 1; u <= n; ++u)
    if (!p[u]) dfs_pre(u);
  for (int u = 1; u <= n; ++u)
    if (!p[u]) dfs(u, 0);
  for (int i = 1; i < m; ++i) printf("%d ", ans[i]);
  printf("%d\n", ans[m]);
}
int main() {
  input();
  solve();
  return 0;
}
