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;
inline int readint() {
  int a;
  cin >> a;
  return a;
}
int toInt(string &s) {
  const static int Mod = 998244353;
  const static long long bas = 29;
  int res = 0;
  for (int i = 0, n = s.length(); i < n; ++i)
    res = (res * bas + s[i] - 'a' + 1) % Mod;
  return res;
}
const int MaxN = 100005;
vector<int> v[MaxN];
map<int, int> *dp[MaxN];
int siz[MaxN], son[MaxN];
void build(int x) {
  siz[x] = 1, son[x] = x;
  for (auto y : v[x]) {
    dp[y] = dp[x] + 1, build(y);
    if (siz[y] >= siz[son[x]]) son[x] = y;
    siz[x] += siz[y];
  }
}
int val[MaxN];
void calc(int x, bool f) {
  if (f)
    dp[x][0][val[x]]++;
  else {
    dp[x][0][val[x]]--;
    if (dp[x][0][val[x]] == 0) dp[x][0].erase(val[x]);
  }
  for (auto y : v[x]) calc(y, f);
}
vector<pair<int, int> > query[MaxN];
int ans[MaxN];
void dfs(int x, bool f) {
  for (auto y : v[x])
    if (y != son[x]) dfs(y, false);
  if (son[x] != x) dfs(son[x], true);
  for (auto y : v[x])
    if (y != son[x]) calc(y, true);
  dp[x][0][val[x]]++;
  for (auto t : query[x]) {
    int dep = t.first;
    int id = t.second;
    ans[id] = dp[x][dep].size();
  }
  if (not f) calc(x, false);
}
map<int, int> cnt[MaxN];
bool root[MaxN];
int main() {
  int n = readint();
  string s;
  int fa;
  for (int i = 1; i <= n; ++i) {
    cin >> s >> fa;
    if (!fa)
      root[i] = true;
    else
      v[fa].push_back(i);
    val[i] = toInt(s);
  }
  for (int i = 1; i <= n; ++i)
    if (root[i]) dp[i] = cnt, build(i);
  int T = readint();
  pair<int, int> p;
  for (int i = 1, a, b; i <= T; ++i) {
    a = readint();
    b = readint();
    if (dp[a] + b >= cnt + MaxN) continue;
    p.first = b;
    p.second = i;
    query[a].push_back(p);
  }
  for (int i = 1; i <= n; ++i)
    if (root[i]) dfs(i, false);
  for (int i = 1; i <= T; ++i) printf("%d\n", ans[i]);
  return 0;
}
