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 N = 200005;
int st[N + N], ft[N + N], vt[N + N], sz[N], dpth[N];
int n, m, par[N], ans[N];
string names[N];
vector<int> adj[N];
int tmt = 1;
vector<pair<int, int> > qs[N];
map<string, int> mp[N];
set<string> ds[N];
void predfs(int x) {
  st[x] = tmt++;
  vt[st[x]] = x;
  for (int y : adj[x]) {
    dpth[y] = dpth[x] + 1;
    predfs(y);
    sz[x] += sz[y];
  }
  sz[x]++;
  ft[x] = tmt++;
}
void dfs(int x, int keep) {
  int mx = -1, bc = -1;
  for (int y : adj[x]) {
    if (sz[y] > mx) {
      mx = sz[y];
      bc = y;
    }
  }
  for (int y : adj[x]) {
    if (y != bc) {
      dfs(y, 0);
    }
  }
  if (bc != -1) {
    dfs(bc, 1);
  }
  for (int y : adj[x]) {
    if (y == bc) continue;
    for (int t = st[y]; t < ft[y]; t++) {
      if (vt[t] == 0) continue;
      mp[dpth[vt[t]]][names[vt[t]]]++;
      if (mp[dpth[vt[t]]][names[vt[t]]] == 1) {
        ds[dpth[vt[t]]].insert(names[vt[t]]);
      }
    }
  }
  mp[dpth[x]][names[x]]++;
  if (mp[dpth[x]][names[x]] == 1) ds[dpth[x]].insert(names[x]);
  for (auto p : qs[x]) {
    int val = p.first + dpth[x], idx = p.second;
    ans[idx] = ds[val].size();
  }
  if (keep) return;
  for (int t = st[x]; t < ft[x]; t++) {
    if (vt[t] == 0) continue;
    mp[dpth[vt[t]]][names[vt[t]]]--;
    if (mp[dpth[vt[t]]][names[vt[t]]] == 0) {
      ds[dpth[vt[t]]].erase(names[vt[t]]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> names[i] >> par[i];
    adj[par[i]].push_back(i);
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    qs[x].push_back({y, i});
  }
  for (int x : adj[0]) {
    dpth[x] = 1;
    predfs(x);
    dfs(x, 0);
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}
