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 double PI = acos(-1);
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
long long Add(long long a, long long b) {
  long long ret = a + b;
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
long long subtract(long long a, long long b) {
  long long ret = a - b;
  if (ret < 0) ret += 1000000007;
  return ret;
}
long long mult(long long a, long long b) {
  long long ret = a * b;
  if (ret >= 1000000007) ret %= 1000000007;
  return ret;
}
long long bigmod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = mult(ret, a);
    b >>= 1;
    a = mult(a, a);
  }
  return ret;
}
long long inverse(long long n) { return bigmod(n, 1000000007 - 2); }
bool base[1000010];
vector<int> primes;
void sieve(int mx) {
  mx += 10;
  int x = sqrt(mx);
  for (int i = 3; i <= x; i += 2)
    if (base[i] == 0)
      for (int j = i * i, k = i << 1; j < mx; j += k) base[j] = 1;
  primes.push_back(2);
  for (int i = 3; i < mx; i += 2)
    if (base[i] == 0) primes.push_back(i);
}
bool checkBit(int n, int i) { return (n & (1 << i)); }
int setBit(int n, int i) { return (n | (1 << i)); }
int resetBit(int n, int i) { return (n & (~(1 << i))); }
int sub[500010], big[500010], ans[500010], par[500010];
vector<int> e[500010];
string s[500010];
vector<pair<int, int> > q[500010];
set<int> cnt[500010];
map<string, int> idx;
void dfs(int u) {
  sub[u] = 1;
  for (int &v : e[u]) {
    dfs(v);
    sub[u] += sub[v];
  }
}
void rem(int u, int d) {
  auto it = cnt[d].find(idx[s[u]]);
  if (it != cnt[d].end()) cnt[d].erase(it);
  for (int &v : e[u])
    if (!big[v]) {
      rem(v, d + 1);
    }
}
void add(int u, int d) {
  cnt[d].insert(idx[s[u]]);
  for (int &v : e[u])
    if (!big[v]) {
      add(v, d + 1);
    }
}
void dfs(int u, int d, int keep) {
  int mx = -1, bigchild = -1;
  for (int &v : e[u]) {
    if (sub[v] > mx) {
      mx = sub[v], bigchild = v;
    }
  }
  for (int &v : e[u])
    if (v != bigchild) {
      dfs(v, d + 1, 0);
    }
  if (bigchild != -1) {
    dfs(bigchild, d + 1, 1);
    big[bigchild] = 1;
  }
  add(u, d);
  for (pair<int, int> &it : q[u]) ans[it.second] = cnt[it.first + d].size();
  if (bigchild != -1) {
    big[bigchild] = 0;
  }
  if (keep == 0) {
    rem(u, d);
  }
}
int main() {
  int n, Q, h, u;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    cin >> s[i];
    if (idx[s[i]] == 0) idx[s[i]] = i;
    scanf("%d", &par[i]);
    e[par[i]].push_back(i);
  }
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    scanf("%d%d", &u, &h);
    q[u].push_back({h, i});
  }
  for (int i = 1; i <= n; ++i)
    if (par[i] == 0) {
      dfs(i);
      dfs(i, 0, 0);
    }
  for (int i = 0; i < Q; ++i) printf("%d\n", ans[i]);
  return 0;
}
