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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MAXN = 2e5 + 100;
int N, Q, a[MAXN];
int hshsz;
map<string, int> hsh;
vector<int> g[MAXN];
int sz[MAXN], depth[MAXN], timer, tin[MAXN], tou[MAXN], vert[MAXN];
void getsz(int v, int p) {
  vert[timer] = v;
  tin[v] = timer++;
  sz[v] = 1;
  for (int u : g[v]) {
    if (u ^ p) {
      depth[u] = depth[v] + 1;
      getsz(u, v);
      sz[v] += sz[u];
    }
  }
  tou[v] = timer - 1;
}
int cnt[MAXN];
map<pair<int, int>, int> mp;
void add(int l, int r, int op) {
  for (int i = l; i <= r; i++) {
    int v = vert[i];
    pair<int, int> state = {a[v], depth[v]};
    if (op == 1) {
      mp[state]++;
      if (mp[state] == 1) cnt[depth[v]]++;
    } else if (op == -1) {
      mp[state]--;
      if (mp[state] == 0) cnt[depth[v]]--;
    }
  }
}
vector<pair<int, int> > queries[MAXN];
int ans[MAXN];
void process_queries(int v) {
  for (auto [dep, index] : queries[v]) {
    ans[index] = cnt[depth[v] + dep];
  }
  queries[v].clear();
}
void dfs(int v, int p, int rem) {
  int mxsz = 0, bigch = -1;
  for (int u : g[v]) {
    if (u ^ p) {
      if (sz[u] > mxsz) {
        mxsz = sz[u], bigch = u;
      }
    }
  }
  for (int u : g[v]) {
    if (u ^ p) {
      if (u != bigch) {
        dfs(u, v, 1);
      }
    }
  }
  if (bigch != -1) {
    dfs(bigch, v, 0);
  }
  add(tin[v], tin[v], +1);
  for (int u : g[v]) {
    if (u ^ p) {
      if (u != bigch) {
        add(tin[u], tou[u], +1);
      }
    }
  }
  process_queries(v);
  if (rem == 1) {
    add(tin[v], tin[v], -1);
    for (int u : g[v]) {
      if (u ^ p) {
        add(tin[u], tou[u], -1);
      }
    }
  }
}
void solve() {
  cin >> N;
  vector<bool> is_root(N, false);
  for (int i = (0); i < (N); i++) {
    string s;
    int p;
    cin >> s >> p;
    if (hsh.count(s))
      a[i] = hsh[s];
    else
      (a[i] = hsh[s] = hshsz++);
    --p;
    if (p == -1)
      is_root[i] = true;
    else
      g[p].push_back(i);
  }
  cin >> Q;
  for (int q = (0); q < (Q); q++) {
    int v, k;
    cin >> v >> k;
    --v;
    queries[v].push_back({k, q});
  }
  for (int v = (0); v < (N); v++) {
    if (is_root[v]) {
      getsz(v, -1);
      dfs(v, -1, 1);
    }
  }
  for (int q = (0); q < (Q); q++) {
    cout << ans[q] << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  solve();
  return 0;
}