Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const long long mod = 1e9 + 7;
const long long maxn = 1e5 + 10;
const long long K = 755;
long long n;
long long p[maxn], h[maxn], par[maxn][22], st[maxn];
vector<long long> g[maxn], ve;
void dfs(long long u) {
  par[u][0] = p[u];
  for (long long i = 1; (1 << i) <= h[u]; i++)
    par[u][i] = par[par[u][i - 1]][i - 1];
  st[u] = 1;
  for (long long i = 0; i < g[u].size(); i++)
    if (g[u][i] != p[u]) {
      h[g[u][i]] = h[u] + 1;
      p[g[u][i]] = u;
      dfs(g[u][i]);
      st[u] += st[g[u][i]];
    }
  ve.push_back(u);
}
long long get_par(long long u, long long he) {
  if (he == 0) return u;
  if (he == 1) return par[u][0];
  long long x = log2(he);
  return get_par(par[u][x], he - (1 << x));
}
vector<long long> bin[maxn];
long long idx[maxn];
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i]) {
      g[p[i]].push_back(i);
      g[i].push_back(p[i]);
    }
  }
  for (long long i = 1; i <= n; i++)
    if (p[i] == 0) dfs(i);
  for (long long i = 0; i < n; i++) {
    idx[ve[i]] = i;
    bin[h[ve[i]]].push_back(i);
  }
  long long q;
  cin >> q;
  while (q--) {
    long long v, k;
    cin >> v >> k;
    if (h[v] < k) {
      cout << 0 << "\n";
      continue;
    }
    long long x = get_par(v, k);
    long long l = idx[x] - st[x] + 1, r = idx[x];
    long long xx =
        lower_bound(bin[h[v]].begin(), bin[h[v]].end(), l) - bin[h[v]].begin();
    long long yy = upper_bound(bin[h[v]].begin(), bin[h[v]].end(), r) -
                   bin[h[v]].begin() - 1;
    cout << yy - xx << "\n";
  }
}