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 INF = 1e9 + 7;
const int N = 1e5 + 7;
const int M = 1e6 + 7;
const long long mod = 1e9 + 7;
int n, c[N];
vector<int> G[N], roots;
map<string, int> id;
int sz[N], son[N], dep[N];
void dfs(int u, int fa) {
  sz[u] = 1;
  dep[u] = dep[fa] + 1;
  for (int v : G[u])
    if (v != fa) {
      dfs(v, u);
      sz[u] += sz[v];
      if (sz[v] > sz[son[u]]) son[u] = v;
    }
}
vector<pair<int, int> > query[N];
int ans[N];
set<int> vis[N];
void update(int u, int fa, bool opt, int Ison) {
  if (opt)
    vis[dep[u]].insert(c[u]);
  else
    vis[dep[u]].clear();
  for (int v : G[u])
    if (v != fa && v != Ison) update(v, u, opt, Ison);
}
void dsu(int u, int fa, int opt) {
  for (int v : G[u])
    if (v != fa && v != son[u]) dsu(v, u, 0);
  if (son[u]) dsu(son[u], u, 1);
  update(u, fa, 1, son[u]);
  for (auto now : query[u]) ans[now.first] = vis[dep[u] + now.second].size();
  if (!opt) update(u, fa, 0, 0);
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n;
  int cnt = 0;
  for (int v = 1; v <= n; v++) {
    string sa;
    int u;
    cin >> sa >> u;
    if (!id[sa]) id[sa] = ++cnt;
    c[v] = id[sa];
    if (u)
      G[u].push_back(v);
    else
      roots.push_back(v);
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int u, k;
    cin >> u >> k;
    query[u].push_back(make_pair(i, k));
  }
  for (int root : roots) {
    dfs(root, 0);
    dsu(root, 0, 0);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
  return 0;
}
