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 = 1e5 + 2;
const int M = 4e5 + 2;
const long long mod = 1e9 + 7;
const int inf = 1e9;
const long long INF = 1e16;
void data() {}
int n, m, p, v, k, ans[N], cn[N], ss[N], h[N], cnt = 0;
string s[N], s1;
vector<int> g[N], root;
vector<pair<int, int> > q[N];
map<string, int> mm;
struct trie {
  int end;
  trie *nxt[10];
  trie() {
    for (int i = 0; i < 10; ++i) nxt[i] = nullptr;
    end = 0;
  }
} * col[N];
void add(int v, int gl) {
  trie *cnt = col[gl];
  for (int i = 0; i < (int)s[v].size(); ++i) {
    if (cnt->nxt[s[v][i] - '0'] == nullptr) {
      cnt->nxt[s[v][i] - '0'] = new trie();
    }
    cnt = cnt->nxt[s[v][i] - '0'];
  }
  cnt->end++;
  if (cnt->end == 1) cn[gl]++;
}
void del(string s, int gl) {
  trie *cnt = col[gl];
  for (int i = 0; i < (int)s.size(); ++i) cnt = cnt->nxt[s[i] - '0'];
  cnt->end--;
  if (cnt->end == 0) cn[gl]--;
}
void dfs_add(int v) {
  add(v, h[v]);
  for (auto to : g[v]) dfs_add(to);
}
void dfs_del(int v) {
  del(s[v], h[v]);
  for (auto to : g[v]) dfs_del(to);
}
void prec(int v, int d = 1) {
  h[v] = d;
  ss[v] = 1;
  for (auto to : g[v]) {
    prec(to, d + 1);
    ss[v] += ss[to];
  }
}
void dfs(int v, bool cl) {
  int mx = 0, mxs = 0;
  for (auto to : g[v])
    if (mxs < ss[to]) mx = to, mxs = ss[to];
  for (auto to : g[v])
    if (to != mx) dfs(to, 1);
  if (mx) dfs(mx, 0);
  for (auto to : g[v])
    if (to != mx) dfs_add(to);
  add(v, h[v]);
  for (auto it : q[v]) ans[it.second] = cn[h[v] + it.first];
  if (cl) dfs_del(v);
}
int main() {
  data();
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) col[i] = new trie();
  for (int i = 1; i <= n; ++i) {
    cin >> s1 >> p;
    if (mm[s1] == 0) mm[s1] = cnt++;
    s[i] = to_string(mm[s1]);
    if (!p) root.push_back(i);
    g[p].push_back(i);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d", &v, &k);
    q[v].push_back(make_pair(k, i));
  }
  for (auto it : root) {
    prec(it);
    dfs(it, 1);
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}
