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 long long MAXN = 100 * 1000 + 10, L = 20;
vector<long long> child[MAXN], roots;
vector<pair<long long, long long> > q[MAXN];
map<long long, long long> mp[MAXN];
long long hei[MAXN], par[L + 5][MAXN], ans[MAXN];
void pre_dfs(long long v, long long p) {
  hei[v] = hei[p] + 1;
  par[0][v] = p;
  for (long long i = 1; i < L; i++) par[i][v] = par[i - 1][par[i - 1][v]];
  for (long long u : child[v]) pre_dfs(u, v);
}
long long find_par(long long v, long long x) {
  for (long long i = L - 1; i >= 0; i--)
    if (x & (1 << i)) v = par[i][v];
  return v;
}
void dfs(long long v) {
  mp[v][hei[v]]++;
  for (long long u : child[v]) {
    dfs(u);
    if ((long long)mp[v].size() < (long long)mp[u].size()) mp[v].swap(mp[u]);
    for (auto p : mp[u]) mp[v][p.first] += p.second;
    mp[u].clear();
  }
  for (pair<long long, long long> p : q[v]) ans[p.first] = mp[v][p.second] - 1;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long p;
    cin >> p;
    if (p)
      child[p - 1].push_back(i);
    else
      roots.push_back(i);
  }
  for (long long r : roots) pre_dfs(r, r);
  long long m;
  cin >> m;
  for (long long i = 0; i < m; i++) {
    long long u, k;
    cin >> u >> k;
    u--;
    long long anc = find_par(u, k);
    if (hei[u] - hei[anc] != k) continue;
    q[anc].push_back(pair<long long, long long>(i, hei[u]));
  }
  for (long long r : roots) dfs(r);
  for (long long i = 0; i < m; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
