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;
string val[100007];
unordered_map<string, int> mp[100007];
vector<int> adj[100007];
vector<pair<int, int> > qq[100007];
int sub[100007], lvl[100007];
int cnt[100007];
int ans[100007];
bool mark[100007];
int st[100007], en[100007], rev[100007];
int Time;
void dfs(int u, int pre) {
  st[u] = ++Time;
  lvl[u] = lvl[pre] + 1;
  rev[Time] = u;
  sub[u]++;
  for (auto v : adj[u]) {
    dfs(v, u);
    sub[u] += sub[v];
  }
  en[u] = Time;
}
void sack(int u, bool keep) {
  int boro = 0, bigchild = -1;
  for (auto v : adj[u])
    if (sub[v] > boro) boro = sub[v], bigchild = v;
  for (auto v : adj[u])
    if (v != bigchild) sack(v, 0);
  if (bigchild != -1) sack(bigchild, 1);
  for (auto v : adj[u])
    if (v != bigchild)
      for (int i = st[v]; i <= en[v]; i++) mp[lvl[rev[i]]][val[rev[i]]]++;
  mp[lvl[u]][val[u]]++;
  for (auto q : qq[u])
    if (q.first < 100007) ans[q.second] = mp[q.first].size();
  if (!keep)
    for (int i = st[u]; i <= en[u]; i++) {
      mp[lvl[rev[i]]][val[rev[i]]]--;
      if (!mp[lvl[rev[i]]][val[rev[i]]]) mp[lvl[rev[i]]].erase(val[rev[i]]);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> val[i] >> p;
    if (p)
      adj[p].push_back(i);
    else
      mark[i] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (mark[i]) dfs(i, i);
  cin >> q;
  for (int i = 0; i < q; i++) {
    int u, k;
    cin >> u >> k;
    qq[u].push_back(pair<int, int>(k + lvl[u], i));
  }
  for (int i = 1; i <= n; i++)
    if (mark[i]) sack(i, 0);
  for (int i = 0; i < q; i++) cout << ans[i] << "\n";
}