#include <bits/stdc++.h>
using namespace std;
int const N = 100001;
int n, q, h[N], en[N], ex[N], cm, fm[N], pa[N][18], lg[N], mxl;
vector<vector<pair<int, int> > > x;
vector<int> tr[N], f;
int t = 0;
void pl(int v, int p = 0, int hi = 0) {
  en[v] = ++t;
  h[v] = hi;
  fm[v] = cm;
  pa[v][0] = p;
  for (int i = 1; i < mxl; ++i)
    if (pa[v][i - 1])
      pa[v][i] = pa[pa[v][i - 1]][i - 1];
    else
      break;
  x[cm].push_back(make_pair(h[v], en[v]));
  for (int i = 0; i < tr[v].size(); ++i)
    if (tr[v][i] != p) pl(tr[v][i], v, hi + 1);
  ex[v] = t + 1;
}
int main() {
  scanf("%d", &n);
  int z = 0, r = 2;
  for (int i = 0; i < n; ++i) {
    int t;
    scanf("%d", &t);
    if (t)
      tr[t].push_back(i + 1);
    else
      f.push_back(i + 1);
    if (i + 1 == r) r <<= 1, ++z;
    lg[i + 1] = z;
  }
  mxl = lg[n] + 1;
  for (int i = 0; i < f.size(); ++i) {
    x.resize(i + 1);
    pl(f[i]);
    sort(x[cm].begin(), x[cm].end());
    ++cm;
  }
  scanf("%d", &q);
  while (q--) {
    int v, l;
    scanf("%d%d", &v, &l);
    if (h[v] < l)
      printf("0 ");
    else {
      int t = h[v];
      while (l) {
        v = pa[v][lg[l]];
        l -= 1 << lg[l];
      }
      int an =
          lower_bound(x[fm[v]].begin(), x[fm[v]].end(), make_pair(t, ex[v])) -
          lower_bound(x[fm[v]].begin(), x[fm[v]].end(), make_pair(t, en[v]));
      printf("%d ", an - 1);
    }
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

Polycarpus got hold of a family relationship tree. The tree describes family relationships of n people, numbered 1 through n. Each person in the tree has no more than one parent.

Let's call person a a 1-ancestor of person b, if a is the parent of b.

Let's call person a a k-ancestor (k > 1) of person b, if person b has a 1-ancestor, and a is a (k - 1)-ancestor of b's 1-ancestor. 

Family relationships don't form cycles in the found tree. In other words, there is no person who is his own ancestor, directly or indirectly (that is, who is an x-ancestor for himself, for some x, x > 0).

Let's call two people x and y (x ≠ y) p-th cousins (p > 0), if there is person z, who is a p-ancestor of x and a p-ancestor of y.

Polycarpus wonders how many counsins and what kinds of them everybody has. He took a piece of paper and wrote m pairs of integers vi, pi. Help him to calculate the number of pi-th cousins that person vi has, for each pair vi, pi.

Input

The first input line contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. The next line contains n space-separated integers r1, r2, ..., rn, where ri (1 ≤ ri ≤ n) is the number of person i's parent or 0, if person i has no parent. It is guaranteed that family relationships don't form cycles.

The third line contains a single number m (1 ≤ m ≤ 105) — the number of family relationship queries Polycarus has. Next m lines contain pairs of space-separated integers. The i-th line contains numbers vi, pi (1 ≤ vi, pi ≤ n).

Output

Print m space-separated integers — the answers to Polycarpus' queries. Print the answers to the queries in the order, in which the queries occur in the input.

Examples

Input

6
0 1 1 0 4 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

0 0 1 0 0 1 1 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int const N = 100001;
int n, q, h[N], en[N], ex[N], cm, fm[N], pa[N][18], lg[N], mxl;
vector<vector<pair<int, int> > > x;
vector<int> tr[N], f;
int t = 0;
void pl(int v, int p = 0, int hi = 0) {
  en[v] = ++t;
  h[v] = hi;
  fm[v] = cm;
  pa[v][0] = p;
  for (int i = 1; i < mxl; ++i)
    if (pa[v][i - 1])
      pa[v][i] = pa[pa[v][i - 1]][i - 1];
    else
      break;
  x[cm].push_back(make_pair(h[v], en[v]));
  for (int i = 0; i < tr[v].size(); ++i)
    if (tr[v][i] != p) pl(tr[v][i], v, hi + 1);
  ex[v] = t + 1;
}
int main() {
  scanf("%d", &n);
  int z = 0, r = 2;
  for (int i = 0; i < n; ++i) {
    int t;
    scanf("%d", &t);
    if (t)
      tr[t].push_back(i + 1);
    else
      f.push_back(i + 1);
    if (i + 1 == r) r <<= 1, ++z;
    lg[i + 1] = z;
  }
  mxl = lg[n] + 1;
  for (int i = 0; i < f.size(); ++i) {
    x.resize(i + 1);
    pl(f[i]);
    sort(x[cm].begin(), x[cm].end());
    ++cm;
  }
  scanf("%d", &q);
  while (q--) {
    int v, l;
    scanf("%d%d", &v, &l);
    if (h[v] < l)
      printf("0 ");
    else {
      int t = h[v];
      while (l) {
        v = pa[v][lg[l]];
        l -= 1 << lg[l];
      }
      int an =
          lower_bound(x[fm[v]].begin(), x[fm[v]].end(), make_pair(t, ex[v])) -
          lower_bound(x[fm[v]].begin(), x[fm[v]].end(), make_pair(t, en[v]));
      printf("%d ", an - 1);
    }
  }
}

```