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 = 100000 + 10;
const int M = 1000000007;
const double PI = atan(1) * 4;
const int oo = 1000000000;
int n, m, T, v[N], ans[N], d[N], sz[N];
vector<int> src;
vector<vector<int> > g;
vector<pair<int, int> > q[N];
unordered_map<string, int> hsh;
char ss[21];
unordered_map<int, unordered_set<int> > mp[N];
unordered_map<int, unordered_set<int> >::iterator it1;
unordered_set<int>::iterator it2;
void dfs(int u) {
  sz[u] = 1;
  int mx = 0, id = -1;
  for (int x, i = 0; i < g[u].size(); ++i) {
    x = g[u][i];
    d[x] = d[u] + 1;
    dfs(x);
    sz[u] += sz[x];
    if (sz[x] > mx) {
      mx = sz[x];
      id = x;
    }
  }
  if (id == -1) {
    mp[u][d[u]].insert(v[u]);
    return;
  }
  mp[u].swap(mp[id]);
  mp[u][d[u]].insert(v[u]);
  for (int x, i = 0; i < g[u].size(); ++i) {
    x = g[u][i];
    if (x == id) continue;
    for (it1 = mp[x].begin(); it1 != mp[x].end(); ++it1) {
      int dth = it1->first;
      unordered_set<int> &st = it1->second;
      for (it2 = st.begin(); it2 != st.end(); ++it2) mp[u][dth].insert(*it2);
    }
  }
  for (int x, i = 0; i < q[u].size(); ++i) {
    x = q[u][i].first + d[u];
    if (mp[u].find(x) != mp[u].end()) ans[q[u][i].second] = mp[u][x].size();
  }
}
inline int get(string s) {
  if (hsh.find(s) != hsh.end()) return hsh[s];
  return hsh[s] = T++;
}
int main() {
  cin >> n;
  g.resize(n);
  for (int x, i = 0; i < n; ++i) {
    scanf(" %s%d", ss, &x);
    v[i] = get(ss);
    if (x)
      g[x - 1].push_back(i);
    else
      src.push_back(i);
  }
  cin >> m;
  for (int a, b, i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    q[a - 1].push_back(pair<int, int>(b, i));
  }
  for (int i = 0; i < src.size(); ++i) dfs(src[i]);
  for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
}
