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;
long long time2 = 0;
vector<long long> p;
vector<vector<long long>> a;
vector<long long> used;
vector<pair<long long, long long>> time1;
vector<vector<long long>> bp;
vector<long long> gl;
vector<vector<long long>> gl1;
void dfs1(long long v) {
  used[v] = 1;
  ++time2;
  time1[v].first = time2;
  for (long long i = 0; i < a[v].size(); ++i) {
    if (!used[a[v][i]]) {
      gl[a[v][i]] = gl[v] + 1;
      dfs1(a[v][i]);
    }
  }
  ++time2;
  time1[v].second = time2;
}
long long parent(long long v, long long k) {
  for (long long i = 24; i >= 0; --i) {
    if (k > (1 << i)) {
      v = bp[v][i];
      k -= (1 << i);
    }
  }
  return p[v];
}
long long binary_search1(long long i, long long x) {
  long long l = -1, r = gl1[i].size();
  while (r - l > 1) {
    long long m = (l + r) / 2;
    if (gl1[i][m] >= x) {
      r = m;
    } else {
      l = m;
    }
  }
  return r;
}
long long binary_search2(long long i, long long x) {
  long long l = -1, r = gl1[i].size();
  while (r - l > 1) {
    long long m = (l + r) / 2;
    if (gl1[i][m] <= x) {
      l = m;
    } else {
      r = m;
    }
  }
  return l;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n;
  p.resize(n + 1);
  p[n] = n;
  a.resize(n);
  used.resize(n);
  time1.resize(n);
  bp.resize(n + 1);
  gl.resize(n);
  gl1.resize(n);
  for (long long i = 0; i < n; ++i) {
    long long tmp;
    cin >> tmp;
    if (tmp == 0) {
      p[i] = n;
    } else {
      p[i] = tmp - 1;
      a[i].push_back(tmp - 1);
      a[tmp - 1].push_back(i);
    }
  }
  for (long long i = 0; i < n + 1; ++i) {
    bp[i].resize(25);
  }
  for (long long i = 0; i < 25; ++i) {
    bp[n][i] = n;
  }
  for (long long i = 0; i < 25; ++i) {
    for (long long j = 0; j < n; ++j) {
      if (i == 0) {
        bp[j][i] = p[j];
      } else {
        bp[j][i] = bp[bp[j][i - 1]][i - 1];
      }
    }
  }
  for (long long i = 0; i < n; ++i) {
    if (p[i] == n) {
      gl[i] = 0;
      dfs1(i);
    }
  }
  for (long long i = 0; i < n; ++i) {
    gl1[gl[i]].push_back(time1[i].first);
  }
  for (long long i = 0; i < n; ++i) {
    sort(gl1[i].begin(), gl1[i].end());
  }
  cin >> m;
  for (long long i = 0; i < m; ++i) {
    long long v, k;
    cin >> v >> k;
    --v;
    v = parent(v, k);
    if (v == n) {
      cout << 0 << " ";
    } else {
      long long l = binary_search1(gl[v] + k, time1[v].first);
      long long r = binary_search2(gl[v] + k, time1[v].second);
      cout << max((long long)0, r - l) << " ";
    }
  }
}