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;
long long int sz[500005], big[500005], ans[500005];
vector<pair<long long int, long long int> > query[500005];
set<string> s[500005];
string nm[100005];
vector<long long int> adj[500005];
void dfs_sz(long long int n, long long int p) {
  for (auto x : adj[n]) {
    if (x == p) continue;
    dfs_sz(x, n);
    sz[n] += sz[x];
  }
  sz[n]++;
}
void add(long long int n, long long int p, long long int x, long long int dis) {
  if (x == 1) {
    s[dis].insert(nm[n]);
  } else {
    s[dis].erase(nm[n]);
  }
  for (auto xx : adj[n]) {
    if (xx == p || big[xx] == 1) continue;
    add(xx, n, x, dis + 1);
  }
}
void dfs(long long int n, long long int p, long long int keep,
         long long int dis) {
  long long int mx = -1, bigg = -1;
  for (auto x : adj[n]) {
    if (x == p) continue;
    if (sz[x] > mx) {
      mx = sz[x];
      bigg = x;
    }
  }
  for (auto x : adj[n]) {
    if (x != bigg && x != p) {
      dfs(x, n, 0, dis + 1);
    }
  }
  if (bigg != -1) {
    dfs(bigg, n, 1, dis + 1);
    big[bigg] = 1;
  }
  add(n, p, 1, dis);
  for (auto d : query[n]) {
    long long int ind = d.second;
    long long int hgt = d.first;
    ans[ind] = s[hgt + dis].size();
    if (n == 1 && hgt == 1) {
    }
  }
  if (bigg != -1) big[bigg] = 0;
  if (!keep) {
    add(n, p, -1, dis);
  }
}
signed main() {
  long long int n, m, vv, h, i, p;
  vector<long long int> src;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) {
    cin >> nm[i];
    scanf("%lld", &p);
    if (p != 0) {
      adj[i].push_back(p);
      adj[p].push_back(i);
    } else {
      src.push_back(i);
    }
  }
  scanf("%lld", &m);
  for (i = 0; i < m; i++) {
    scanf("%lld %lld", &vv, &h);
    query[vv].push_back({h, i});
  }
  for (auto x : src) {
    dfs_sz(x, -1);
    dfs(x, -1, 0, 1);
  }
  for (i = 0; i < m; i++) {
    printf("%lld\n", ans[i]);
  }
}
