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 maxn = (int)1e5 + 100;
const int P = (int)1e6 + 7;
const int mod = (int)1e9 + 7;
int n, q, lvl[maxn], ans[maxn], d[maxn], num, a[maxn];
vector<int> g[maxn], root;
vector<pair<int, int> > Q[maxn];
map<int, int> id;
map<int, set<int> > cnt[maxn];
void add(int v, int to) {
  for (auto x : cnt[to])
    for (auto val : x.second) cnt[v][x.first].insert(val);
  cnt[to].clear();
}
void dfs(int v, int p = 0) {
  lvl[v] = lvl[p] + 1;
  d[v] = 1;
  int ch = -1;
  for (int to : g[v]) {
    dfs(to, v);
    d[v] += d[to];
    if (ch == -1 || d[ch] < d[to]) ch = to;
  }
  if (ch != -1) swap(cnt[v], cnt[ch]);
  cnt[v][lvl[v]].insert(a[v]);
  for (int to : g[v]) {
    if (to != ch) add(v, to);
  }
  for (auto x : Q[v]) {
    ans[x.second] = ((int)(cnt[v][lvl[v] + x.first]).size());
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    string second;
    int p;
    cin >> second >> p;
    if (!p) root.push_back(i);
    g[p].push_back(i);
    int cur = 0;
    for (auto x : second) cur = cur * P + x;
    if (!id[cur]) id[cur] = ++num;
    a[i] = id[cur];
  }
  cin >> q;
  for (int i = (1); i <= (q); ++i) {
    int v, k;
    cin >> v >> k;
    Q[v].push_back(make_pair(k, i));
  }
  for (int v : root) dfs(v);
  for (int i = (1); i <= (q); ++i) cout << ans[i] << endl;
}
