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;
const int N = 100005, M = N;
struct adjacency {
  int head[N], nxt[N], to[N], n, ne;
  void init(int n) {
    this->n = n;
    ne = 0;
    memset(head, -1, n * sizeof(head[0]));
  }
  void add_edge(int f, int t) {
    to[ne] = t;
    nxt[ne] = head[f];
    head[f] = ne++;
  }
  void add_bi_edge(int u, int v) {
    add_edge(u, v);
    add_edge(v, u);
  }
} adj, query;
string names[N];
int parent[N], k[M], ans[M], sz[N];
void calc_size(int u) {
  sz[u] = 1;
  for (int v, e = adj.head[u]; ~e && (v = adj.to[e], 1); e = adj.nxt[e]) {
    calc_size(v);
    sz[u] += sz[v];
  }
}
deque<unordered_set<string> > small_to_large(int u) {
  deque<unordered_set<string> > ret;
  int large = -1;
  for (int v, e = adj.head[u]; ~e && (v = adj.to[e], 1); e = adj.nxt[e]) {
    if (sz[v] * 2 > sz[u])
      large = v;
    else
      small_to_large(v);
  }
  if (~large) ret = small_to_large(large);
  ret.push_front(unordered_set<string>({names[u]}));
  for (int v, e = adj.head[u]; ~e && (v = adj.to[e], 1); e = adj.nxt[e]) {
    if (sz[v] * 2 > sz[u]) continue;
    queue<int> q;
    q.push(v);
    int level = 1;
    while (!q.empty()) {
      int q_sz = q.size();
      while (q_sz--) {
        v = q.front();
        q.pop();
        if (level >= ret.size()) ret.resize(level + 1);
        ret[level].insert(names[v]);
        for (int w, e = adj.head[v]; ~e && (w = adj.to[e], 1); e = adj.nxt[e])
          q.push(w);
      }
      ++level;
    }
  }
  for (int i, e = query.head[u]; ~e && (i = query.to[e], 1); e = query.nxt[e]) {
    if (k[i] < ret.size()) ans[i] = ret[k[i]].size();
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int n, m;
  cin >> n;
  adj.init(n);
  query.init(n);
  vector<int> roots;
  for (int i = 0; i < n; ++i) {
    cin >> names[i] >> parent[i];
    if (parent[i])
      adj.add_edge(parent[i] - 1, i);
    else
      roots.push_back(i);
  }
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int v;
    cin >> v >> k[i];
    query.add_edge(v - 1, i);
  }
  for (int root : roots) {
    calc_size(root);
    small_to_large(root);
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << '\n';
  return 0;
}