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 nm[100005];
vector<int> v[100005];
bool vis[100005];
vector<pair<int, int> > names[100005];
int level[100005], f[100005], l[100005], tim, mxlevel;
map<string, int> mp;
void dfs(int node, int lev) {
  vis[node] = true;
  level[node] = lev;
  mxlevel = max(mxlevel, lev);
  f[node] = tim;
  names[lev].push_back(pair<int, int>(f[node], mp[nm[node]]));
  for (int i = 0; i < (v[node].size()); i++) {
    if (vis[v[node][i]]) continue;
    tim++;
    dfs(v[node][i], lev + 1);
  }
  l[node] = tim;
  tim++;
}
vector<pair<int, int> > qu[100005];
vector<pair<int, int> > tmp[100005];
int ans[100005];
void add(vector<int> &bit, int val, int idx) {
  while (idx < bit.size()) {
    bit[idx] += val;
    idx += ((idx) & (-(idx)));
  }
}
int sum(vector<int> &bit, int idx) {
  int ans = 0;
  while (idx) {
    ans += bit[idx];
    idx -= ((idx) & (-(idx)));
  }
  return ans;
}
void fans(int lev) {
  int i;
  vector<int> bit(names[lev].size() + 1, 0);
  for ((i) = 0; (i) < (bit.size()); (i)++) tmp[i].clear();
  int ll, r, j;
  vector<pair<int, int> > it1, it2;
  for ((i) = 0; (i) < (qu[lev].size()); (i)++) {
    ll = lower_bound(names[lev].begin(), names[lev].end(),
                     pair<int, int>(f[qu[lev][i].first], -1)) -
         names[lev].begin() + 1;
    r = lower_bound(names[lev].begin(), names[lev].end(),
                    pair<int, int>(l[qu[lev][i].first], -1)) -
        names[lev].begin();
    if (ll >= ((int)(bit.size())) || r <= 0 || r < ll) {
      ans[qu[lev][i].second] = 0;
      continue;
    }
    ll = min(ll, ((int)(bit.size())) - 1);
    r = min(r, ((int)(bit.size())) - 1);
    tmp[r].push_back(pair<int, int>(ll, qu[lev][i].second));
  }
  map<int, int> pr;
  for (i = 1; i < bit.size(); i++) {
    add(bit, 1, i);
    if (pr.find(names[lev][i - 1].second) != pr.end()) {
      add(bit, -1, pr[names[lev][i - 1].second]);
    }
    pr[names[lev][i - 1].second] = i;
    for ((j) = 0; (j) < (tmp[i].size()); (j)++) {
      ans[tmp[i][j].second] = sum(bit, i) - sum(bit, tmp[i][j].first - 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, n, m, val;
  vector<int> anc;
  string nam;
  cin >> n;
  for ((i) = 0; (i) < (n); (i)++) {
    cin >> nm[i] >> val;
    if (val == 0)
      anc.push_back(i);
    else
      v[val - 1].push_back(i);
    mp[nm[i]] = 0;
  }
  i = 1;
  for (map<string, int>::iterator it = mp.begin(); it != mp.end(); it++)
    it->second = i++;
  for ((i) = 0; (i) < (anc.size()); (i)++) dfs(anc[i], 0);
  cin >> m;
  int v, k;
  for ((i) = 0; (i) < (m); (i)++) {
    cin >> v >> k;
    v--;
    if (level[v] + k > mxlevel)
      ans[i] = 0;
    else {
      qu[level[v] + k].push_back(pair<int, int>(v, i));
    }
  }
  for ((i) = 0; (i) < (mxlevel + 1); (i)++) fans(i);
  for ((i) = 0; (i) < (m); (i)++) cout << ans[i] << endl;
  return 0;
}