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 = 1e5 + 5;
int n;
string S[N];
int col[N], p, k, m;
vector<int> roots, adj[N];
vector<pair<int, int> > Queries[N];
int got[N];
int sz[N], in[N], out[N], ord[N << 1], lvl[N], ans[N], timer;
map<int, int> cnt[N];
void init() {
  cin >> n;
  vector<string> ord;
  for (int i = 1; i <= n; i++) {
    cin >> S[i] >> p;
    ord.push_back(S[i]);
    if (p)
      adj[p].push_back(i);
    else
      roots.push_back(i);
  }
  sort(ord.begin(), ord.end());
  ord.resize(unique(ord.begin(), ord.end()) - ord.begin());
  for (int i = 1; i <= n; i++) {
    col[i] = lower_bound(ord.begin(), ord.end(), S[i]) - ord.begin() + 1;
  }
}
void getQueries() {
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> p >> k;
    Queries[p].push_back(make_pair(i, k));
  }
}
void pre(int u, int p) {
  sz[u] = 1;
  in[u] = timer;
  ord[timer++] = u;
  for (auto v : adj[u])
    if (v != p) {
      lvl[v] = lvl[u] + 1;
      pre(v, u);
      sz[u] += sz[v];
    }
  out[u] = timer++;
}
void dfs(int u, int p, bool keep) {
  int mx = -1, bigChild = -1;
  for (auto v : adj[u])
    if (v != p && sz[v] > mx) mx = sz[v], bigChild = v;
  for (auto v : adj[u])
    if (v != p && v != bigChild) dfs(v, u, 0);
  if (~bigChild) dfs(bigChild, u, 1);
  for (auto v : adj[u])
    if (v != p && v != bigChild)
      for (int x = in[v]; x < out[v]; x++)
        if (ord[x]) {
          if (cnt[col[ord[x]]][lvl[ord[x]]] == 0) got[lvl[ord[x]]]++;
          cnt[col[ord[x]]][lvl[ord[x]]]++;
        }
  cnt[col[u]][lvl[u]]++;
  got[lvl[u]]++;
  for (auto v : Queries[u])
    if (v.second + lvl[u] <= n) ans[v.first] = got[lvl[u] + v.second];
  if (!keep) {
    for (int x = in[u]; x < out[u]; x++)
      if (ord[x]) {
        if (cnt[col[ord[x]]][lvl[ord[x]]] == 1) got[lvl[ord[x]]]--;
        cnt[col[ord[x]]][lvl[ord[x]]]--;
      }
  }
}
void solve() {
  for (auto v : roots) {
    pre(v, v);
    dfs(v, v, 0);
  }
}
void print() {
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  init();
  getQueries();
  solve();
  print();
}